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This mock test of Test: Unit Digit- 1 for UPSC helps you for every UPSC entrance exam.
This contains 15 Multiple Choice Questions for UPSC Test: Unit Digit- 1 (mcq) to study with solutions a complete question bank.
The solved questions answers in this Test: Unit Digit- 1 quiz give you a good mix of easy questions and tough questions. UPSC
students definitely take this Test: Unit Digit- 1 exercise for a better result in the exam. You can find other Test: Unit Digit- 1 extra questions,
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QUESTION: 1

What will be the units digit, when the product of the first 10 natural numbers is divided by 100?

Solution:

QUESTION: 2

Find the product of 57 x 61 x 39 x 53.

Solution:

QUESTION: 3

Find the units digit of the product of all the prime numbers between 1 and 13^{13}.

Solution:

QUESTION: 4

What is the rightmost non-zero digit of 90^{42}?

Solution:

QUESTION: 5

Find the units digit of 5^{3n} + 9^{5m}, where m and n are positive integers

- m is an odd integer
- n is an even integer

Solution:

QUESTION: 6

If x = 3^{21} and y = 6^{55}, 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 = 3^{21} and y = 6^{55}, 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 = 3^{21}

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 3^{21} = 3

Next, let’s find the unit digit of y:

y = 6^{55}

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 31467^{32 }× 97645^{23} × (32168^{5} + 8652)^{479}?

Solution:

**Step 1: Question statement and Inferences**

We need to find the units digit of all the terms in the expression 31467^{32 }× 97645^{23} × (32168^{5} + 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: 31467^{32}

The unit digit of this term will be determined by 7^{32}

32 = 4*8

--> 32 is of the form 4m

--> 7’s power is of the form 4m

--> The units digit of 7^{32 }will be 1

--> The units digit of 31467^{32} is 1

Considering the second term: 97645^{23}

The units digit of this term will be determined by 5^{23}

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

--> The units digit of 5^{23 }is 5

--> The units digit of 97645^{23 }is 5

Considering the third term: 32168^{5}

The units digit of this term will be determined by 8^{5}

5 = 4*1 +1

--> 5 is of the form 4m +1

--> 8’s power is of the form 4m +1

--> The units digit of 8^{5 }will be 8

--> The units digit of 32168^{5} is 8

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

**Step 3: Calculating the final answer**

Units digit of 31467^{32} × 97645^{23} × (32168^{5} + 8652)^{479 }= 1 × 5 (8 + 2)^{479} = 1 × 5 (0)^{479} = 1 × 5 × 0 = 0

**Answer: Option (A) **

QUESTION: 8

Find the rightmost non-zero digit of the number 345637300^{3725}.

Solution:

**Step 1: Question statement and Inferences**

We are given the number 345637300^{3725}, 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:

345637300^{3725} = (345673 * 100)^{3725}

= 345673^{3725} * 100^{3725}

Now, we know that the rightmost non-zero digit of the number will come from the expression 345673^{3725}.

Also, the unit digit of the expression 345673^{3725} = the unit digit of 3^{3725}

Thus, we have to find the unit digit of 3^{3725}.

**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 3^{3725} = the unit digit of 3^{4m + 1} = 3

**Step 3: Calculating the final answer**

So, the rightmost non-zero digit of the number 345637300^{3725} = 3.

**Answer: Option (B)**

QUESTION: 9

If p is a positive integer, what is the units digit of Z, if Z = (104^{4p + 1}) * (277^{p + 1}) * (93^{p + 2}) * (309^{6p}) ?

Solution:

**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)**

QUESTION: 10

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

(1) q = 2p – 11

(2) q^{2} – 10q + 9 = 0

Solution:

__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:

q^{2} – 10q + 9 = 0

q^{2} – 9q -q + 9 = 0

(q – 9) (q – 1) = 0

Thus, q= 1, 9

Let’s consider both the values one by one:

- If q = 1

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

7^{1+23 }= 7^{24}

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 7^{24} = 1

2. If q = 9

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

79^{+23 }= 7^{32}

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 7^{32} = 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)**

QUESTION: 11

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

Solution:

90 = 10 x 9

Clearly, 653*xy* is divisible by 10, so *y* = 0

Now, 653*x*0 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 7^{105} ?

Solution:

Unit digit in 7^{105} = Unit digit in [ (7^{4})^{26} x 7 ]

But, unit digit in (7^{4})^{26} = 1

Unit digit in 7^{105} = (1 x 7) = 7

QUESTION: 14

Which of the following numbers will completely divide (4^{61} + 4^{62} + 4^{63} + 4^{64}) ?

Solution:

(4^{61} + 4^{62} + 4^{63} + 4^{64}) = 4^{61} x (1 + 4 + 4^{2} + 4^{3}) = 4^{61} x 85

= 4^{60} x (4 x 85)

= (4^{60} 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:** *(a ^{2} - b^{2}) = (a + b)(a - b)*]

= (200 x 12)

= 2400.

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