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# Test: Unit Digit- 2

## 10 Questions MCQ Test Quantitative Aptitude for GMAT | Test: Unit Digit- 2

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This mock test of Test: Unit Digit- 2 for UPSC helps you for every UPSC entrance exam. This contains 10 Multiple Choice Questions for UPSC Test: Unit Digit- 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Unit Digit- 2 quiz give you a good mix of easy questions and tough questions. UPSC students definitely take this Test: Unit Digit- 2 exercise for a better result in the exam. You can find other Test: Unit Digit- 2 extra questions, long questions & short questions for UPSC on EduRev as well by searching above.
QUESTION: 1

### For an integer n = 216 * 313 * 55 * 75 * 113, what is the value of the first non-zero digit from the right?

Solution:

Given:

• An integer n whose prime-factorized form is 216 * 313 * 55 * 75 * 113

To find: The first non-zero digit from the right

Approach:

1. To answer the question, we first need to understand why, in the first place, the rightmost digits of n will be 0.
• The answer lies in the presence of both 2 and 5 in n
2. So, we’ll calculate the number of zeros in n
3. Then, we will separate the prime numbers of n (along with their powers) that do not contribute to these zeros. The first non-zero digit from the right will be equal to the units digit of the product of these separated prime numbers (along with their powers)
4. Using the cyclicity of these separated prime numbers, we’ll get to the answer

Working Out:

• Calculating the number of zeros in n
• The zeros result from the product of 2 and 5
• Each zero requires one 2 and one 5
• In n, there are sixteen 2s and five 5s
• So, 5 is the limiting factor in the making of zeros
• The number of zeros will be equal to the number of 5s in n
• So, there will be five zeros in n
• All the five 5s and five of the sixteen 2s in n will be used up by the zeros

• Separating the prime numbers that do not contribute to the zeros
• As noted above, all the five 5s and five 2s of n are used up by zeros
• So, the prime numbers of n (along with powers) that do not contribute to the zeros are:
• 211
• 313
• 75
• 113

• So, the right-most non-zero digit of n is actually the units digit of the product 211*313*75*113
• The units digit of this product will be obtained by multiplying the units digits of:
• 211
• 313
• 75
• 113

• Using the cyclicity of the separated prime numbers to get the answer
• Finding Units digit of 211
• 211  = 24*2+3
• So, units digit of 211 = units digit of 23 = 8

• Finding Units digit of 313
• 313  = 34*3+1
• So, units digit of 313 = units digit of 31 = 3

• Finding Units digit of 75
• 75  = 74+1
• So, units digit of 75 = units digit of 71 = 7

• Finding Units digit of 113
• Units digit = 1

• Finding units digit of the product 211*313*75*113
• = Units digit of (8*3*7*1)
• = 8

Looking at the answer choices, we see that the correct answer is Option E

QUESTION: 2

### What is the units digit of (a +b)2 – (a-b)2, where a and b are non-negative integers?        (1) The difference between any two consecutive multiples of a is 5        (2) b when multiplied by any even integer results in the same units digit, not necessarily equal to units digit of b.

Solution:

Steps 1 & 2: Understand Question and Draw Inferences Step 3: Analyze Statement 1 independently

1. The difference between any two consecutive multiples of a is 5

• Let the consecutive multiples of a be a*n and a*(n+1)
• So, a(n+1) – an = 5
• an + a – an = 5
• a = 5

If a = 5, 4ab = 20b. So, irrespective of the value of b, the units digit(20b) = 0.

Step 4: Analyze Statement 2 independently

2.  b when multiplied by any even integer results in the same units digit, not necessarily equal to units digit of b.

• Following cases are possible:

• Units Digit(b) = 0.
• 0 when multiplied with any integer will have the Units Digit as 0. In this case the units digit(4ab) = 0.

• Units Digit(b) = 5.
• 5 when multiplied with any even integer will result in units digit of 0. So, Units Digit(4ab) = 0.

As we have a unique answer of Units Digit(4ab), this statement is sufficient to answer.

Step 5: Analyze Both Statements Together (if needed)

As we have a unique answer from steps 3 and 4, this step is not required.

QUESTION: 3

### What is the units digit of zx, where x and z are positive integers?      (1)   z when divided by 100 has its hundredths digit as 5      (2)  The product of z2 and z3 has the same units digit as z2.

Solution:

Steps 1 & 2: Understand Question and Draw Inferences

• x, z are integers > 0

To Find: Units Digit (zx)

Step 3: Analyze Statement 1 independently

1. z when divided by 100 has its hundredths digit as 5

• Let z = abcdef, where f is the units digit.
• So, z100=abcd.ef

The hundredths digit of z100

•  = f = Units Digit (z) = 5

Now, a number with its Units Digit = 5, raised to any power will always have its Units Digit as 5.

So, Units Digit (zx) = 5

(Note: Even if you took z to have, say, 8 digits – a number like z = abcdefgh, where a to h are digits – you would still get the units digit of z (the units digit is h in this case) as 5. Therefore, the answer would still be the same)

Step 4: Analyze Statement 2 independently

2. The product of z2 and z3 has the same units digit as z2.

• Product of z2 and z3 = (z2)(z3) = z2+3 = z5
• Units Digit (z5) = Units Digit (z2). This equation will be satisfied in the following cases:
• Case 1: If the units digit of z2 has a cyclicity of 3
• However, we know that none of the digits have a cyclicity of 3
• Therefore, this case is rejected
• Case 2: If the units digit of z2 has a cyclicity of 1
• The digits that satisfy this case are: { 0, 1, 5, 6}
• So, Units Digit(z) = { 0, 1, 5, 6}

As we do not have a unique value of Units Digit of(z), this statement is insufficient to answer.

Step 5: Analyze Both Statements Together (if needed)

As we have a unique answer from step-3, this step is not required.

QUESTION: 4

For any positive number n, the function #n represents the value of the number n rounded to the nearest integer. If k is a positive number, what is the units digit of #k?

(1)  #(10k) = 10k

(2)  #(100k) is 10300.

Solution:

Steps 1 & 2: Understand Question and Draw Inferences

Given:

• Let k = n + a.bcde. . . , where n is a multiple of 10 and a, b, c, d, e etc. denote the units, tenths, hundredths, thousandths, ten-thousandths digits of k respectively.

• Example: Say k is equal to 123.789. We can also write this value as 120 + 3.789. This form is comparable to n + a.bcde . . .

• So, n denotes the value of all the digits at the tens, hundreds, thousands and higher place values (if k = 123.789, n = 120; if k = 3456.987, n = 3450; if k = 43789.34, n = 43780 and so on)

• We have split k into n and a.bcde. . . because the question doesn’t indicate whether the magnitude of k is in tens or hundreds or thousands etc. (that is, is k a number like 12.345 or like 123.789 or like 3456.987 etc.) This knowledge is not important either because all the action is happening at the units digit and the digits to the right of the decimal point. So, we’ve simply expressed the value of all the digits at the tens, hundreds, thousands and higher place values in a single term n

• #k =
• n + a, if b < 5
• n + (a+1) if b ≥ 5

To find:

• The units digit of #k
• Since n is a multiple of 10, from the expression of #k, we can say that the units digit of #k =
• a, if b < 5
• a + 1, if b ≥ 5

Thus, to answer the question, we need to know the value of a and whether b < 5 or not

Step 3: Analyze Statement 1 independently

(1)  #(10k) = 10k

• 10k = 10n + ab.cde. . .
• Since a is now the tens digit and b is the units digit, we can write

10k = 10n + 10a + b.cde. . .

10k = 10(n+a) + b.cde. . .

• Applying the definition of the function #n, we can write:
• #(10k) =
• 10(n+a) + b if c < 5
• 10(n+a) + (b+1) if c ≥ 5
• We are given that #(10k) = 10k.
• #(10k) = 10(n+a) + b.cde . . .

• If c < 5, then by substituting the value of #(10k) in the above equation, we get:
• 10(n+a) + b = 10(n+a) + b.cde. . .
• b = b + 0.cde. . .
• This implies, c = d = e . . .= 0
• Thus, k is a number of the form n + a.b, where b is the sole digit after the decimal point

• If  c ≥ 5, then by substituting the value of #(10k) in the above equation, we get:
• 10(n+a) + (b+1) = 10(n+a) + b.cde. . .
• b  +1 = b + 0.cde . . .
• 1 = 0.cde . .  .
• For no values of digits c, d, e . . . will the above equation be satisfied.
• Therefore, this case is not possible

• Thus, from Statement 1, we conclude that k is a number of the form n + a.b, where b is the sole digit after the decimal point
• However, we still don’t know the value of a or whether b is less than 5 or not.

So, Statement 1 alone is not sufficient.

Step 4: Analyze Statement 2 independently

(2) #(100k) is 10300.

• 100k = 100n + abc.de. . .
• 100k = (100n + 100a + 10b) + c.de. . .

• Applying the definition of the function #n, we can write:
• #(100k) =
• (100n + 100a + 10b) + c if d < 5
• (100n + 100a + 10b) + (c+1) if d ≥ 5

• We are given that #(100k) = 10300
• If d < 5, then we can write
• (100n + 100a + 10b) + c = 10300
• Remember that in the above expression, a, b and c are digits (therefore, lie between 0 and 9, inclusive) whereas n is a multiple of 10.
• Comparing the units digits on both sides of the equation, we get: c = 0
• Comparing the tens digits on both sides of the equation, we get: b = 0
• Comparing the hundreds digits on both sides of the equation, we get: a = 3
• Since b < 5 in this case, the units digit of #k = a = 3

• If d ≥ 5, then we can write
• (100n + 100a + 10b) + (c+1) = 10300
• Comparing the units digits on both sides of the equation, we get: c + 1 = 0
• So, c = 9 and 1 is carried over to tens place
• Comparing the tens digits on both sides of the equation, we get: b + 1 = 0 (Note, we’re writing b+1 and not b due to the carry-over)
• So, b = 9 and 1 is carried over to hundreds place
• Comparing the hundreds digits on both sides of the equation, we get: a + 1 = 3
• So, a = 2
• Since b > 5 in this case, the units digit of #k = a  +1 = 3

• Thus, we see that both possible values of #(100k) lead to the same value of the units digit of #k
• Thus, Statement 2 is sufficient to find a unique answer to the question.

Step 5: Analyze Both Statements Together (if needed)

Since we’ve already arrived at a unique answer in Step 4, this step is not required

QUESTION: 5

Arrange the following terms in the increasing order of their units digits:

I. 75

II. 86

III. 123

Solution:

Given: 3 terms: 75, 86 and 123

To find: The correct increasing order of the units digits of the 3 terms

Approach:

1. We will first find the units digit of each term by using the cyclicity of units digits
2. Then, we will arrange the units digits in increasing order.

Working Out:

• Finding the units digit of Term I - 75
• We know that 7 has a cyclicity of 4
• Now, 5 = 4 + 1
• So, the units digit of 75 will be the same as the units digit of 71 = 7
• So, units digit of Term I is 7

• Finding the units digit of Term II - 86
• We know that 8 has a cyclicity of 4
• Now, 6 = 4 + 2
• So, the units digit of 86 will be the same as the units digit of 82 = 64
• So, units digit of Term II is 4

• Finding the units digit of Term III - 123
• The units digit of 123 will come from 23 = 8
• So, units digit of Term III is 8

• Arranging the units digits in increasing order, we get: 4 < 7 < 8
• That is, II < I < III
• Looking at the answer choices, we see that the correct answer is Option D
QUESTION: 6

For any positive number x, the function [x] denotes the greatest integer less than or equal to x. For example,  = 1, [1.367] = 1 and [1.999] = 1.

If k is a positive integer such that k2 is divisible by 45 and 80, what is the units digit of Solution:

Given:

• Positive integer k (so, k > 0)
• k2 is divisible by 45
• k2 is divisible by 80
• The function [x]

To find: The units digit of Approach:

1. To find the units digit of [x], we need to know or at least have some idea of the value of x. So, in order to answer the question, we need to find the value or at least a clue about the value of the number 1. This number contains k3, which is the cube of a positive integer. However, the given information is about k2. Like in all questions that involve divisibility information about different powers of an integer, we’ll work out information about k3 as under:
• First use the information given about k2 to infer about the prime-factorized form of k.
• Then, by cubing the prime-factorized expression of k, get an expression for k3

Working Out:

• Inferring the prime-factorized expression for k from the given information
• Let k = P1a * P2b *P3c *P4d  . . .where P1, P2 etc. are prime numbers and a, b . . . are non-negative integers
• Therefore, k2 = P12a * P22b *P32c *P42d  . . .
• We are given that k2 is divisible by 45
• 45 = 32*5
• This means, P1 = 3 and 2a ≥ 2
• That is, a ≥ 1
• And, P2 = 5 and 2b ≥ 1
• That is, • Since b is an integer, minimum possible value of b = 1
• Therefore, k = (31*51)(3x*5y* P3c *P4d  . . .) where x and y are non-negative integers
• Note: In the above expression, we’ve taken 31 and 51 outside the remaining expression in order to emphasize the fact that k definitely does contain 31 and 51, whether or not it contains higher powers of 3 and 5
• So far, we’ve inferred that: k = (31*51)(3x*5y* P3c *P4d  . . .)
• So, k2 = (32*52)(32x*52y* P32c *P42d  . . .)
• We are also given that k2 is divisible by 80
• 80 = 24*5
• This means, P3 = 2 and 2c ≥ 4
• That is, c ≥ 2
• So, k = (22*31*51)(2z*3x*5y* P4d  . . .), where x, y, z, d etc. are non-negative integers
• Inferring the expression for k3
• Using the expression for k inferred above, we can write:
• k3 = (26*33*53)(23z*33x*53y* P43d  . . .),
• Drawing inferences about the value of • 4000 = 25*53
• Looking at the expression for k3 above, we see that k3 will be divisible by 4000

• So, is an integer
• The value of this integer will be (21*33)(23z*33x*53y* P43d  . . .)
• So, = (21*33)(23z*33x*53y* P43d  . . .)
• If y ≥ 1, then the units digit of the right hand side expression will be 0 (because the units digit of 2*5 is 0. Note that the above expression has at least one 2)
• But if y = 0, then the units digit will depend on the power of 2, 3 and the value and powers of any other prime numbers that are present in k
•

• Thus, we are not able to determine a unique value of the units digit of • Looking at the answer choices, we see that the correct answer is Option E

QUESTION: 7

n is a positive integer that lies between 100 and 200, exclusive, and has no digits repeated. Is the units digit of n equal to 5?

(1)  n is divisible by 9 and the 2-digit number formed by inverting the units and tens digit of n is a prime number.

(2)  The difference between the units digit and the tens digit of n is the same as the difference between the tens digit and the hundreds digit of n and is equal to 2.

Solution:

Steps 1 & 2: Understand Question and Draw Inferences

• n is an integer with no repeated digits such that 100 < n < 200
• From the above inequality, it is clear that Hundreds digit(n) = 1
• Let the number n = 1bc, where b and c are the tens and units digits of n respectively.

To Find: Is Units Digit(n) = 5?

• That is, is c = 5?

Step 3: Analyze Statement 1 independently

1. n is divisible by 9 and the 2-digit number formed by inverting the units and tens digit of n is a prime number.
• The number cb is a prime number
• Also, since n is divisible by 9, the sum of the digits of n must be divisible by 9
• Therefore, 1 + b + c must be divisible by 9
• So, 1 + b + c = { 9, 18}.
• It cannot be equal to 27 or more as the maximum possible sum of b + c = 9 +8 = 17 (Remember that the digits of n are distinct. So, if one digit is 9, the maximum value of the other digit can be 8).
• So, following cases are possible:
• Either b + c = 8. Possible combinations of two distinct positive integers adding up to 8 can be:

• (b, c) = (2, 6). In this case cb = 62, which is not a prime number. So, this case is not possible.
• (b, c) = (3, 5). In this case, cb = 53, which is a prime number. So, this is a possible case.
• (b, c) = (5, 3). In this case cb = 35, which is not a prime number. So, this case is not possible
• (b, c) = (6, 2). In this case cb = 26, which is not a prime number. So, this case is not possible.

• Or b + c = 17. In this case, (b, c) = (8, 9) in any order. Following cases are possible:

• If (b, c) = (8, 9), then cb = 98, which is not a prime number. So, this case is not possible
• If (b, c) = (9, 8), then cb = 89, which is a prime number. So, this is a possible case.

Hence, the possible values of n = { 135, 198}. So, the units digit of n = {5, 8}

As we do not have a unique value of units digit(n), the statement is insufficient to answer.

Step 4: Analyze Statement 2 independently

2.   The difference between the units digit and the tens digit of n is the same as the difference between the tens digit and the hundreds digit of n and is equal to 2.

• We’ve already inferred in Steps 1 and 2 that Hundreds digit(n) = 1
• So, Tens Digit(n) = 1 + 2 = 3
• And, Units Digit(n) = 3 + 2 = 5

Thus, the units digit(n) = 5

As we have a unique value of units digit(n) , this statement is sufficient to answer.

Step 5: Analyze Both Statements Together (if needed)

As we have a unique value from step 4, this step is not required.

QUESTION: 8

If p and q are positive integers, what is the remainder when (27)12p + (3)6q is divided by 5?

(1) p is an odd prime number

(2) q is divisible by 10

Solution:

Steps 1 & 2: Understand Question and Draw Inferences

Given: Positive integers p and q

To find:

• The remainder when (27)12p + (3)6q is divided by 5
• The remainder when an expression is divided by 5 is determined by the units digit of the expression  • The units digit of the expression is obtained from the sum of the units digits of (27)12p and (3)6q
1. Units digit of 2712p = Units digit of 712p = Units digit of (74)3p = Units digit of 13p = 1
2. Units digit of (3)6q = Units digit of (32)3q = Units digit of (9)3q
• If q is even, units digit of (9)3q  = units digit of 9even = 1
• If q is odd, units digit of (9)3q = units digit of 9odd = 9

• Thus, the units digit of the expression is equal to
1. 1 + 1 = 2 (if q is even)
2. 1 + 9 = 0 (if q is odd)

• Correspondingly, the remainder when the expression is divided by 5 is equal to
1. 2 (if q is even)
2. 0 (if q is odd)

So, to answer the question, we need to know if q is even or odd.

Step 3: Analyze Statement 1 independently

(1) p is an odd prime number

Statement 1 gives no information about q’s even-odd nature.

So it’s not sufficient to answer the question

Step 4: Analyze Statement 2 independently

(2) q is divisible by 10?

• So, q is of the form 10k, where k is a positive integer
• Thus, q is even

Since we now know the even-odd nature of q, we can answer the question. Statement 2 alone is sufficient.

Step 5: Analyze Both Statements Together (if needed)

Since we’ve already arrived at a unique answer in Step 4, this step is not required

QUESTION: 9

If x and y are distinct positive integers and x+y is even, what is the remainder when (x+y)a is divided by 10, where a is a positive integer?

(1) Units digit of y is 6

(2) (xy)a is divisible by 10.

Solution:

Steps 1 & 2: Understand Question and Draw Inferences

• x, y are distinct integers > 0 such that x + y = even
• This means, Either both x andy  are even
• Or both x andy are odd
• a is an integer > 0

To Find: the value of r in (x+y)a=10k+r, where k is the quotient obtained when (x+y)a  is divided by 10 and r is the remainder; so,  0 ≤ r < 10

• Now, when a number is divided by 10, the remainder  is equal to the units digit of that number.
• So, r= units digit of (x+y)a

Step 3: Analyze Statement 1 independently

1. Units digit of y is 6

• It does not tell us anything about the units digit of x as well as about the value of a.

Step 4: Analyze Statement 2 independently

2.   (xy)a is divisible by 10.

• As (xy)a is divisible by 10, the units digit of (xy)a = 0
• So, the units digit of xy = 0. Two cases are possible:
• Units digit of (x, y) = { 5, even number) in any order. However in this case the number with 5 as its units digit will be odd and the other number will be even. However, we’ve deduced in Steps 1 and 2 that x and y have the same even-odd nature. So, this case is not possible as it contradicts the given information (that the sum x + y is even).
• Units digit of (x, y) = (0, even number) in any order. In this case x and y are both even. So, this case is possible.
• However since we do not have a unique value of units digit of both x and y and we do not know the value of a, we cannot find a unique value of the units digit of (x+y)a

Step 5: Analyze Both Statements Together (if needed)

1. From Statement 1, we know that Units digit of y = 6
2. From Statement 2, we inferred that Units digit of (x, y) = (0, even number) in any order

Combining both the statements, we can say that units digit (x) = 0 and units digit(y) = 6

So, units digit of (x+y) = 6. Now do we need the value of a to find out the units digit of (x+y)?

We know a number with units digit of 6 raised to any power always results in units digit of 6.

So, Units Digit ((6a)) = 6.

Thus r = Units Digit ((6a)) = 6.

QUESTION: 10

If a and b are positive integers, is the sum a + b divisible by 4?

1. When the sum 23a + 25b is divided by 10, the remainder is 8
2. When 22b is divided by 10, the remainder is 8
Solution:

Steps 1 & 2: Understand Question and Draw Inferences

Given: Integers a, b > 0

To find: If a + b is divisible by 4?

Step 3: Analyze Statement 1 independently

Statement 1 says that ‘When the sum 23a + 25b is divided by 10, the remainder is 8’

• Now, we know that when an integer is divided by 10, the remainder is equal to the units digit of that integer
• For example, when 25 is divided by 10, the remainder is 5 (the units digit of 25)
• So, from Statement 1, we can infer that the units digit of the sum 23a + 25b is 8
• Now, the units digit of 25b will always be 5, for all possible values of the positive integer b
• Therefore, the units digit of the term 23a must be 3
• Now, the units digit of 23a comes from 3a
• For the units digit to be 3, a must be of the form 4k + 1, where k is a non-negative integer
• Thus, from Statement 1, we have inferred that a is of the form 4k + 1. However, since we do not yet know about the value of b, we cannot comment if the sum a + b is divisible by 4 or not.
• So, Statement 1 is not sufficient.

Step 4: Analyze Statement 2 independently

Statement 2 says that ‘When 22b is divided by 10, the remainder is 8’

• We know that when an integer is divided by 10, the remainder is equal to the units digit of that integer
• So, from Statement II, we can infer that the units digit of 22b is 8
• Now, the units digit of 22b comes from 2b
• For the units digit to be 8, b must be of the form 4m + 3, where m is a non-negative integer

• Thus, from Statement 2, we have inferred that b is of the form 4m + 3. However, since we do not know about the value of a, we cannot comment if the sum a + b is divisible by 4 or not.
• So, Statement 2 is not sufficient.

Step 5: Analyze Both Statements Together (if needed)

• From Statement 1, we inferred that a is of the form 4k + 1
• From Statement 2, we inferred that b is of the form 4m + 3
• So, combining the 2 statements, we observe that the sum a + b is of the form 4k + 4m + 4
• Therefore the sum a + b is indeed divisible by 4
• Since we’ve arrived at a unique answer to the question, the two statements together are sufficient.