Test: Unit Digit- 1 - UPSC MCQ

1x2x3x4x5x6x7x8x9x10 so clearly they have 2 zeroes 
one from 10 and other from 2x5
so multiply then 3x4x6x7x8x9 will give 8 as unit digit

1. Any power of 90 can be split into 9 to that power, times 10 to that power.
    90ⁿ = 9ⁿ × 10ⁿ

2. The factor of 10 to the 42nd power just adds forty-two zeros on the end, so it doesn’t affect the last non-zero digit.

3. We only need the last digit of 9 raised to the 42nd power.

4. Look at the pattern of 9’s last digit:
   – 9¹ ends in 9
   – 9² ends in 1
   – 9³ ends in 9
   – 9⁴ ends in 1
   …and it repeats every two powers (odd exponents give 9, even exponents give 1).

5. Since 42 is even, 9⁴² ends in 1.

Answer: 1

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

3^4m+1 = 3

3^4m+2 = 9  

3^4m+3 = 7

3^4m = 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.

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

 Answer: Option (A)

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.

3^{4m + 1} = 3

3^{4m + 2} = 9  

3^{4m + 3} = 7

3^4m = 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 3^{4m + 1} = 3

Step 3: Calculating the final answer

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

Answer: Option (B)

Step 1: Question statement and Inferences

We are given that Z = (104^{4p + 1}) * (277^{p + 1}) * (93^{p + 2}) * (309^6p). 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 = (4^{4p + 1}) * (7^{p + 1}) * (3^{p + 2}) * (9^6p) 

Step 2: Finding required values

Z = (4^{4p + 1}) * (7^{p + 1}) * (3^{p + 2}) * (9^6p) 

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

Unit digit of 4^{4p + 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 4^{4p + 1} = 4      (Since 4p + 1 is an odd number and every odd power of 4 has the unit digit as 4)

Unit digit of 9^6p:

Every second power of 9 has the same unit digit.

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

So, unit digit of 9^6p = 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 3^p+2 and 7^p+1 since we don’t know the value of p. However, the product of these numbers can be further solved as follows:

7^{p + 1} * 3^{p + 2} = 7^{p + 1} * 3^{p + 1} * 3

                      = (7*3)^{p + 1} * (3)           (Since a^m * b^m = (ab)^m )

                      = (21)^{p + 1} * (3)

Now, we know that the unit digit of the expression 21^p+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.  

Answer: Option (A)

Steps 1 & 2: Understand Question and Draw Inferences

We are given that X = 6^p + 7^q+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 6^p and 7^q+23.

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

Now, we know that the unit digit of 6 raised to any integer power is 6. So, the unit digit of 6^p 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 = 6^p + 7^q+23 = (Unit Digit of 6^p) + (Unit digit of 7^q+23)

    = 6 + (Unit digit of 7^q+23)

 So, the unit digit of 7^q+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 7^q+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 7^q+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 7^q+23 as 1. As derived in the first step, the unit digit of X

= 6 + (Unit digit of 7^q+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.

Answer: Option (B)

1. For divisibility by 10, the last digit y must be 0.

2. Now number is 6 5 3 x 0. For divisibility by 9, the sum of digits 6+5+3+x+0 = 14+x must be a multiple of 9.
   ⇒ 14+x ≡ 0 mod 9
   ⇒ x ≡ 4 mod 9
   ⇒ x = 4.

3. Therefore x + y = 4 + 0 = 4.

Answer: 4 (option c)

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

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

(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.

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.