The sum of the digits of an odd integer n is 35, and the pro

Question

The sum of the digits of an odd integer n is 35, and the product of tens and units digits of n is 36.

What is the remainder when n is divided by 12?

A. 1
B. 5
C. 7
D. 9
E. 11

Edskore · Answer

Nice Problem Solving question — it cleverly layers two constraints (digit sum and digit product) before asking a Remainders question. Let me walk through it step by step.

Key concept being tested: Digit properties combined with Remainders (specifically using divisibility by 3 and by 4 separately to find the remainder mod 12).

Step 1 — Identify the tens and units digits.
The product of the tens digit (T) and units digit (U) is 36. The single-digit factor pairs of 36 are: (4, 9), (9, 4), and (6, 6).

Step 2 — Apply the "odd integer" constraint.
Since n is odd, its units digit must be odd. From our pairs: (4, 9) → units = 9 ✓; (9, 4) → units = 4 ✗; (6, 6) → units = 6 ✗. So T = 4, U = 9.

Common trap: Many students test all pairs without filtering for "odd integer" first, which wastes time and causes errors.

Step 3 — Find n mod 3.
Divisibility rule: n ≡ (sum of digits) mod 3. Sum of all digits = 35. Since 35 = 33 + 2, we get n ≡ 2 (mod 3).

Step 4 — Find n mod 4.
Divisibility rule: n ≡ (last two digits) mod 4. Last two digits form "49". 49 = 48 + 1, so n ≡ 1 (mod 4).

Step 5 — Combine using the Chinese Remainder Theorem (or by inspection).
We need a number that is ≡ 1 (mod 4) AND ≡ 2 (mod 3). Testing candidates from {1, 5, 9, 13...} (i.e., 4k + 1): 1 mod 3 = 1 ✗; 5 mod 3 = 2 ✓.
Therefore, n ≡ 5 (mod 12).

Answer: B (5)

Takeaway: Whenever a remainder mod 12 question appears, immediately split it into mod 3 (digit sum rule) and mod 4 (last-two-digits rule) — combining the results is almost always faster than direct computation.

Dereno · Answer

We don’t know the number of digits of the number, but their sum is 35.

If the number is a five digit one, say abcde.

Then, sum = a+b+c+d+e = 35

Given that the digit is a ODD number.

The product of tens and units digit is 36.

d*e = 36

12*3 OR 4*9 OR 6*6 are the possibilities.

Digits should be within (0-9), so (12*3) cannot qualify.

(6*6) not suitable, as it’s even digits.

Only case is (49) and not (94) , as the latter is an even number.

So, the number is abc49.

a+b+c+4+9 = 35

Thus, a+b+c = 22

What’s the remainder when the number is divided by 12 ?

Divisibility rule of 12 should satisfy both the divisibility rule of 4 and 3.

For 3, the sum of digits is divisible by 3.

For 4, the last two digits should be divisible by 4.

35 when divided by 3, leaves a remainder 2. Therefore , number = 3p+2

49 when divided by 4, leaves a remainder 1. Therefore, number = 4K+1

4K+1 = 3p+2

4K = 3p+1

At k = 1, and p=1 , both give the same value of 5.

K increments in the coefficients of p =3, and p increments in the coefficients of k =4.

At K=1+3 = 4 and at p= 1+4=5 , the values coincide again. The number is 17.

At k = 4+3 =7 and at p = 5+4 = 9, the values coincide again. The number is 29.

The pattern here is  FIRST NUMBER WHICH MATCHES + LCM (divisors)z

= 5 + LCM (3,4)z

= 5+12z

At z=0, we get 5,

At z=1 , we get 17

So, the remainder when divided by 12 is  5.

Option B

The sum of the digits of an odd integer n is 35, and the pro

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GMAT Problem Solving (PS) Questions

The sum of the digits of an odd integer n is 35, and the pro

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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My Rewards