Remainders – MBA CET questions with solutions

Approach: Remainder method: add (divisor − remainder).

Solution:
1056 ÷ 23: 23 × 45 = 1035, remainder = 1056 − 1035 = 21.
Least number to add = 23 − 21 = 2.
Check: 1058 = 23 × 46.

Cetking shortcut: Nearest multiple above: 23 × 46 = 1058 → 1058 − 1056 = 2.

Answer: A

Approach: Remainder with a divisor that divides the original divisor.

Solution:
N = 256k + 73.
256 = 16 × 16, so 256k leaves remainder 0 when divided by 16.
So remainder = 73 mod 16; 73 = 16 × 4 + 9 → remainder 9.

Cetking shortcut: If the new divisor is a factor of the old one, just divide the old remainder by the new divisor.

Answer: C

Approach: Units digit of a product depends only on the units digits of the factors.

Solution:
Units digits: 5, 2, 7, 4.
5 × 2 = 10 → units digit 0.
Anything × (number ending in 0) ends in 0.
Units digit of 15 × 22 × 37 × 44 = 0.

Why the other options are wrong:
A, D, E (2, 4, 8) – come from ignoring the 5 × 2 pair.
B (5) – 5 × odd ends in 5, but here 5 meets an even number.

Cetking shortcut: A factor ending in 5 together with any even factor always gives units digit 0.

Answer: C

Approach: Remainders multiply: rem(N²) = rem(r²).

Solution:
N = 5k + 3, so N leaves remainder 3.
N² leaves the same remainder as 3² = 9 when divided by 5.
9 = 5 × 1 + 4 ⇒ remainder 4.
Check: N = 8, N² = 64 = 5 × 12 + 4 ✓.

Why the other options are wrong:
C (3) – assumes the remainder stays the same.
A, B, E – arithmetic slips.

Cetking shortcut: Pick the smallest such number (3): 3² = 9 → remainder 4.

Answer: D

Approach: Remainder of a sum = remainder of the sum of individual remainders.

Solution:
Sum = 73 + 75 + 78 + 57 + 197 = 480.
34 × 14 = 476.
480 − 476 = 4.

Cetking shortcut: Remainders by 34: 5 + 7 + 10 + 23 + 27 = 72 → 72 − 68 = 4.

Answer: B

Approach: Number = LCM(5, 6, 7, 8) × k + 3, then force divisibility by 9.

Solution:
LCM(5, 6, 7, 8) = 840 → N = 840k + 3.
840 = 9 × 93 + 3, so N ≡ 3k + 3 (mod 9).
Need 3k + 3 divisible by 9 → k + 1 divisible by 3 → smallest k = 2.
N = 1680 + 3 = 1683 (digit sum 18 ✓ divisible by 9).

Cetking shortcut: Options must be (multiple of 840) + 3 and have digit sum divisible by 9 — only 1683 (1680 + 3, digit sum 18).

Answer: A

Approach: Dividend = Divisor × Quotient + Remainder.

Solution:
30879 = 67 × d + 327.
67d = 30879 − 327 = 30552.
d = 30552/67 = 456 (remainder 327 < 456 ✓).

Why the other options are wrong:
A 476, C 466, D 444, E 457: 67 × each + 327 ≠ 30879 (e.g. 67 × 457 + 327 = 30946).

Cetking shortcut: Units digit: 67 × d must end in 2 (…879 − …327) ⇒ 7 × d ends in 2 ⇒ d ends in 6: options 476, 456, 466 — then 30552/67 = 456.

Answer: B

Approach: aⁿ − bⁿ is divisible by (a + b) when n is even.

Solution:
n = 176 is even ⇒ 354¹⁷⁶ − 346¹⁷⁶ is divisible by 354 + 346 = 700.
Remainder = 0.

Why the other options are wrong:
A, C, D, E: non-zero remainders are impossible since 700 divides the expression exactly.

Cetking shortcut: Spot 354 + 346 = 700 and the even power ⇒ remainder 0 instantly.

Answer: B

Approach: Successive division — reconstruct the number from the last quotient.

Solution:
Let the last quotient be k. Work from the innermost division outwards.
Third division (by 7) leaves 4: the quotient before it = 7k + 4.
Second division (by 4) leaves 1: the quotient before it = 4(7k + 4) + 1 = 28k + 17.
First division (by 3) leaves 2: N = 3(28k + 17) + 2 = 84k + 53.
Note 3 × 4 × 7 = 84, so N = 84k + 53 ⇒ remainder on dividing by 84 = 53.
Check with k = 0: N = 53 → 53 ÷ 3 = 17 r 2; 17 ÷ 4 = 4 r 1; 4 ÷ 7 = 0 r 4. ✓

Cetking shortcut: Remainder by the product of divisors = r1 + d1·r2 + d1·d2·r3 = 2 + 3×1 + 12×4 = 53.

Answer: C

Approach: N = LCM(16, 18, 20, 25)·k + 4, then make N divisible by 7.

Solution: 16 = 2⁴, 18 = 2·3², 20 = 2²·5, 25 = 5² → LCM = 2⁴·3²·5² = 3600.
N = 3600k + 4. 3600 ÷ 7 leaves remainder 2, 4 ÷ 7 leaves 4 → N ≡ 2k + 4 (mod 7).
2k + 4 ≡ 0 (mod 7) → 2k ≡ 3 → k = 5 (2·5 + 4 = 14 ✔).
N = 3600 × 5 + 4 = 18,004.

Cetking shortcut: Every option must be 4 more than a multiple of 3600 → only 18004 fits (18000 + 4); check 18004 ÷ 7 = 2572 ✔.

Answer: D

Approach: Find a small power of 3 that leaves remainder 1 when divided by 11.

Solution:
3⁵ = 243 = 11 × 22 + 1 → 3⁵ ≡ 1 (mod 11).
333 = 5 × 66 + 3 → 3³³³ = (3⁵)⁶⁶ × 3³ ≡ 1 × 27 (mod 11).
27 = 2 × 11 + 5 → remainder 5.

Cetking shortcut: Fermat: 3¹⁰ ≡ 1 (mod 11), but spotting 3⁵ = 243 ≡ 1 is quicker; then only 333 mod 5 = 3 matters.

Answer: E

Approach: Number = LCM × k + remainder; then apply the divisibility-by-7 condition.

Solution:
16 = 2⁴, 18 = 2 × 3², 20 = 2² × 5, 25 = 5² → LCM = 2⁴ × 3² × 5² = 3600.
Number = 3600k + 4.
3600 = 7 × 514 + 2 → 3600k + 4 ≡ 2k + 4 (mod 7). Need 2k + 4 ≡ 0 → 2k ≡ 3 → k ≡ 5 (mod 7).
Smallest k = 5 → 18000 + 4 = 18004 = 7 × 2572 ✓

Cetking shortcut: Only options ending in 4 qualify (18004, 17004); 17004 − 4 = 17000 is not a multiple of 3600, so 18004.

Answer: B

Approach: Same remainder 4 → number = LCM × k + 4; then impose divisibility by 11.

Solution:
LCM(9, 10, 12, 25) = 2² × 3² × 5² = 900 → x = 900k + 4 < 5000 → k = 0 to 5.
900 ≡ 9 (mod 11) → 9k + 4 ≡ 0 (mod 11) → 9k ≡ 7 → k ≡ 2 (mod 11).
Only k = 2 → x = 1804 (= 11 × 164 ✓).
Teams of 12: 1804 ÷ 12 = 150 remainder 4 → maximum 150 teams.

Cetking shortcut: List 4, 904, 1804, 2704, 3604, 4504 and test divisibility by 11 (alternating digit sums): only 1804.

Answer: A

Approach: Replace each factor by its remainder mod 18.

Solution:
1719 = 18 × 95 + 9 → 9
1721 → 11, 1723 → 13, 1725 → 15, 1727 → 17
Product ≡ 9 × (11 × 13 × 15 × 17) (mod 18)
11 × 13 × 15 × 17 is odd, and 9 × odd ≡ 9 (mod 18)
Remainder = 9

Cetking shortcut: One factor leaves 9 and all others are odd: 9 × odd always leaves 9 on division by 18.

Answer: A