# Test: Unit Digit- 1

## 15 Questions MCQ Test Quantitative Reasoning for GMAT | Test: Unit Digit- 1

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Attempt Test: Unit Digit- 1 | 15 questions in 20 minutes | Mock test for UPSC preparation | Free important questions MCQ to study Quantitative Reasoning for GMAT for UPSC Exam | Download free PDF with solutions
QUESTION: 1

Solution:
QUESTION: 2

Solution:
QUESTION: 3

### Find the units digit of the product of all the prime numbers between 1 and 1313.

Solution:
QUESTION: 4

What is the rightmost non-zero digit of  9042?

Solution:
QUESTION: 5

Find the units digit of 53n + 95m, where m and n are positive integers

1. m is an odd integer
2. n is an even integer
Solution:
QUESTION: 6

If x = 321 and y = 655, what is the remainder when xy is divided by 5?

Solution:

The correct option is Option B.

Step 1: Question statement and Inferences

We are given that x = 321 and y = 655, and we have to find the remainder when xy is divided by 5.

When a number is divided by 5, the remainder can be easily calculated if we know the units digit of the number. For example:

If 34 is divided by 5, the remainder is 4. And if 77 is divided by 5, the remainder is 2.

So, here we have to find the unit digit of the product xy, and then we will be able to find the remainder if xy is divided by 5.

Now, the unit digit of the product xy will depend on the unit digits of the individual numbers x and y.

Step 2: Finding required values

Let’s first find the unit digit of x:

x = 321

Cyclicity of 3 is 4

34m+1 = 3

34m+2 = 9

34m+3 = 7

34m = 1

21 = 4 * 5 + 1

4 * 5 + 1 = 4m + 1

Thus, the unit digit of 321 = 3

Next, let’s find the unit digit of y:

y = 655

Cyclicity of 6 is 1

That is, the unit digit of every power of 6 is 6 only.

Therefore, the unit digit of y is 6

Step 3: Calculating the final answer

The unit digit of the product xy = the product of the unit digits of the numbers x and y

= 3 * 6 = 18

Thus, the unit digit of the expression xy is 8

Thus, the remainder when xy is divided by 5 will be 3.

QUESTION: 7

What is the units digit of 3146732 × 9764523 × (321685 + 8652)479?

Solution:

Step 1: Question statement and Inferences

We need to find the units digit of all the terms in the expression 3146732 × 9764523 × (321685 + 8652)479

Step 2: Finding required values

Given:

We Know,

2 → 2, 4, 8, 6 (Corresponding powers: 4m +1,4m+2, 4m +3, 4m)

5 → 5: Every power will have the same last digit.

7 → 7, 9, 3, 1 (Corresponding powers: 4m +1,4m+2, 4m +3, 4m)

8 → 8, 4, 2, 6 (Corresponding powers: 4m +1,4m+2, 4m +3, 4m)

Considering the first term: 3146732

The unit digit of this term will be determined by 732

32 = 4*8

-->  32 is of the form 4m

-->  7’s power is of the form 4m

-->  The units digit of 732 will be 1

-->  The units digit of 3146732 is 1

Considering the second term: 9764523

The units digit of this term will be determined by 523

The units digit of 5 raised to the power anything is 5.

-->  The units digit of 523 is 5

-->  The units digit of 9764523 is 5

Considering the third term: 321685

The units digit of this term will be determined by 85

5 = 4*1 +1

-->  5 is of the form 4m +1

-->  8’s power is of the form 4m +1

-->  The units digit of 85 will be 8

-->  The units digit of 321685 is 8

The units digit of the last term, 8652, is 2

Step 3: Calculating the final answer

Units digit of 3146732 × 9764523 × (321685 + 8652)479 = 1 × 5 (8 + 2)479 = 1 × 5 (0)479  = 1 × 5 × 0 = 0

QUESTION: 8

Find the rightmost non-zero digit of the number 3456373003725

Solution:

Step 1: Question statement and Inferences

We are given the number 3456373003725, and we have to find the rightmost non-zero digit of this number. We know that the rightmost digit is the unit digit of a number.

Now, the expression can be written as follows:

3456373003725 = (345673 * 100)3725

= 3456733725 * 1003725

Now, we know that the rightmost non-zero digit of the number will come from the expression 3456733725.

Also, the unit digit of the expression 3456733725 = the unit digit of 33725

Thus, we have to find the unit digit of 33725.

Step 2: Finding required values

We know that every 4th power of 3 has the same unit digit and cycles of power of 3 are 3, 9, 7, and 1.

34m + 1 = 3

34m + 2 = 9

34m + 3 = 7

34m = 1

Now, 3725 = 3700 + 25

= 4*k + 4*6 + 1                         (Since every number which is a multiple of 100 is a multiple of 4)

So, 3725 = 4m + 1, where m is some positive integer

Thus, the unit digit of 33725 = the unit digit of 34m + 1 = 3

Step 3: Calculating the final answer

So, the rightmost non-zero digit of the number 3456373003725 = 3.

QUESTION: 9

If p is a positive integer, what is the units digit of Z, if Z = (1044p + 1) * (277p + 1) * (93p + 2) * (3096p) ?

Solution:

Step 1: Question statement and Inferences

We are given that Z = (1044p + 1) * (277p + 1) * (93p + 2) * (3096p). We have to find the units digit of Z.

Here we can say that:

The unit digit of Z = The units digit of the product of the unit digits of the given numbers

Now, we also know that the unit digit of any power of a number depends only on the unit digit of the number. Thus, we can write the expression as:

Z = (44p + 1) * (7p + 1) * (3p + 2) * (96p

Step 2: Finding required values

Z = (44p + 1) * (7p + 1) * (3p + 2) * (96p

Next, let’s find the unit digit of the individual expressions:

Unit digit of 44p + 1:

Every second power of 4 has the same unit digit.

Cycles of powers of 4 are 4, 6, 4, 6 …

So, unit digit of 44p + 1 = 4      (Since 4p + 1 is an odd number and every odd power of 4 has the unit digit as 4)

Unit digit of 96p:

Every second power of 9 has the same unit digit.

Cycles of powers of 9 are 9, 1, 9, 1 …

So, unit digit of 96p = 1      (Since 6p is an even number and every even power of 9 has the unit digit 1)

Now, the cyclicity of the numbers 3 and 7 is 4. So, we can’t decide the unit digit of the expression 3p+2 and 7p+1 since we don’t know the value of p. However, the product of these numbers can be further solved as follows:

7p + 1 * 3p + 2 = 7p + 1 * 3p + 1 * 3

= (7*3)p + 1 * (3)           (Since am * bm = (ab)m )

= (21)p + 1 * (3)

Now, we know that the unit digit of the expression 21p+1 will always be 1 since any power of 1 always gives a unit digit 1.

Thus, the unit digit of (21)p + 1 * (3) = 1 * 3 = 3

Step 3: Calculating the final answer

Now, let’s plug in all the values in the expression for Z.

Z = 4 * 3 * 1 = 12

So, the unit digit of Z will be 2.

QUESTION: 10

If p and q are positive integers and X = 6p + 7q+23, what is the units digit of X?

(1) q = 2p – 11

(2) q2 – 10q + 9 = 0

Solution:

Steps 1 & 2: Understand Question and Draw Inferences

We are given that X = 6p + 7q+23, and we have to find the unit digit of X. The numbers p and q both are positive integers. Now, the unit digit of X will be the sum of the unit digits of 6p and 7q+23.

So, we have to find the unit digit of 6p and 7q+23.

Now, we know that the unit digit of 6 raised to any integer power is 6. So, the unit digit of 6p is 6.

And the cyclicity of 7 is 4 i.e. every 4th power of 7 has the same unit digit and cycles of power of 7 are 7, 9, 3, and 1.

So, the unit digit of the expression:

X = 6p + 7q+23 = (Unit Digit of 6p) + (Unit digit of 7q+23)

= 6 + (Unit digit of 7q+23)

So, the unit digit of 7q+23 will depend on the value of q. We have to find the value of q to determine the unit digit of X.

Step 3: Analyze Statement 1

Statement 1 says: q = 2p – 11

However, since we don’t know the value of p, we can’t determine the value of q.

Hence, statement I is not sufficient to answer the question: What is the unit digit of X?

Step 4: Analyze Statement 2

Statement 2 says:

q2 – 10q + 9 = 0

q2 – 9q -q + 9 = 0

(q – 9) (q – 1) = 0

Thus, q= 1, 9

Let’s consider both the values one by one:

1. If q = 1

In this case, the expression 7q+23 becomes:

71+23 = 724

Since the cyclicity of 7 is 4 i.e. every 4th power of 7 has the same unit digit and cycles of power of 7 are 7, 9, 3, and 1.

And,

24 = 4*6

So, the unit digit of 724 = 1

2.  If q = 9

In this case, the expression 7q+23 becomes:

79+23 = 732

Since the cyclicity of 7 is 4 i.e. every 4th power of 7 has the same unit digit and cycles of power of 7 are 7, 9, 3, and 1.

And,

32 = 4*8

So, the unit digit of 732 = 1

So, in both the cases we get the unit digit of 7q+23 as 1. As derived in the first step, the unit digit of X

= 6 + (Unit digit of 7q+23) = 6 + 1 = 7

So, statement (2) alone is sufficient to answer the question: What is the unit digit of X?

Step 5: Analyze Both Statements Together (if needed)

Since statement (2) alone is sufficient to answer the question, we don’t need to perform this step.

QUESTION: 11

If the number 653 xy is divisible by 90, then (x + y) = ?

Solution:

90 = 10 x 9
Clearly, 653xy is divisible by 10, so y = 0
Now, 653x0 is divisible by 9.
So, (6 + 5 + 3 + x + 0) = (14 + x) is divisible by 9. So, x = 4.
Hence, (x + y) = (4 + 0) = 4.

QUESTION: 12

3897 x 999 = ?

Solution:

3897 x 999= 3897 x (1000 - 1)
= 3897 x 1000 - 3897 x 1
= 3897000 - 3897
= 3893103.

QUESTION: 13

What is the unit digit in 7105 ?

Solution:

Unit digit in 7105 = Unit digit in [ (74)26 x 7 ]
But, unit digit in (74)26 = 1 Unit digit in 7105 = (1 x 7) = 7

QUESTION: 14

Which of the following numbers will completely divide (461 + 462 + 463 + 464) ?

Solution:

(461 + 462 + 463 + 464) = 461 x (1 + 4 + 42 + 43) = 461 x 85
= 460 x (4 x 85)
= (460 x 340), which is divisible by 10.

QUESTION: 15

106 x 106 - 94 x 94 = ?

Solution:

106 x 106 - 94 x 94= (106)2 - (94)2
= (106 + 94)(106 - 94)
[Ref: (a2 - b2) = (a + b)(a - b)]
= (200 x 12)
= 2400. Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code