IPMAT Unit Digit PYQs

2 solved Unit Digit previous year questions (PYQs) from past year papers — attempt each and check the answer.

Q1:ipmat indore 2023QA › Unit DigitEasyMCQ · MCQ

Let

p

be a positive integer such that the unit digit of

p ^ 3

4

. What are the possible unit digits of

(p + 3) ^ 3

A3
B1, 7, 9
C4, 7
D1, 3, 7

Pick an option to attempt

The Setup: We are given a positive integer

p

such that the unit digit of

p^3

is exactly

4

. We need to determine the possible unit digits of the expression

(p+3)^3

. Step 1: Determine the unit digit of

p

. We evaluate the cubes of all base digits from

0

9

to see which produces a unit digit of

4

0^3 = 0, 1^3 = 1, 2^3 = 8, 3^3 = 27, 4^3 = 64, 5^3 = 125, 6^3 = 216, 7^3 = 343, 8^3 = 512, 9^3 = 729

. The only digit whose cube ends in

4

4

itself. Therefore, the unit digit of

p

must be

4

. Step 2: Calculate the unit digit of

(p+3)^3

. Since

p \equiv 4 \pmod{10}

, we substitute this into the target expression:

p + 3 \equiv 4 + 3 \equiv 7 \pmod{10}

Now, cube this resulting unit digit:

7^3 = 343

The unit digit of

343

3

. Thus, the only possible unit digit for

(p+3)^3

3

. Final Answer: 3

Q2:ipmat indore 2021QA › Unit DigitMediumMCQ · MCQ

The unit digit in

(743)^{85} - (525)^{37} + (987)^{96}

is ________

Pick an option to attempt

The Setup: Unit digit mechanics operate on a strict modulo-10 cyclicity. We don't care about the massive numbers; we only care about the final digit of the base and the remainder of the exponent when divided by its cycle length. Step 1: Analyze the first term:

743^{85}

. The unit digit depends entirely on

3^{85}

. The powers of

3

cycle in a 4-step loop (

3

9

7

1

). Divide the exponent

85

4

to find the remainder.

85=4\times21+1

Since the remainder is

1

, it hits the first step of the cycle. The unit digit is

3^1=3

. Step 2: Analyze the second term:

525^{37}

. This is a freebie. Any positive integer power of a number ending in

5

will always end in

5

. The unit digit is

5

. Step 3: Analyze the third term:

987^{96}

. The unit digit depends on

7^{96}

. The powers of

7

also cycle in a 4-step loop (

7

9

3

1

). Divide

96

4

96=4\times24+0

Since it divides perfectly (remainder

0

), it lands on the final step of the cycle. The unit digit is

1

. Step 4: Combine the unit digits.

3-5+1=-1

Since we can't output a negative unit digit in a standard integer, we simulate borrowing a

10

from the next placeholder column over (just like elementary school subtraction).

10-1=9

Final Answer: 9

Unit Digit — PYPs

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