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A positive integer n leaves a remainder of 4 when divided by 7 and 28 Jun 2026, 04:27
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D
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35% (medium)

Question Stats:

78% (02:30) correct 22% (02:34) wrong based on 156 sessions

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A positive integer n leaves a remainder of 4 when divided by 7 and a remainder of 5 when divided by 9. What is the remainder when 3n+8 is divided by 63?

A. 17
B. 20
C. 26
D. 41
E. 62
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D
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A positive integer n leaves a remainder of 4 when divided by 7 and 28 Jun 2026, 06:00
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A positive integer n leaves a remainder of 4 when divided by 7 and a remainder of 5 when divided by 9. What is the remainder when 3n+8 is divided by 63?

A. 17
B. 20
C. 26
D. 41
E. 62

Numbers that leave a remainder of 5 when divided by 9 are:

5, 14, 23, 32, 41, ...

Now check which of these leaves a remainder of 4 when divided by 7:

32 divided by 7 leaves remainder 4.

So we can take n = 32.

Then:

3n + 8 = 3 * 32 + 8 = 104

104 divided by 63 leaves remainder 41.

Answer: D.
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A positive integer n leaves a remainder of 4 when divided by 7 and 30 Jul 2026, 23:53
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Is there any way to solve this using equations? Trial and error of all possible numbers may not work if the number is much bigger right?
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A positive integer n leaves a remainder of 4 when divided by 7 and 31 Jul 2026, 02:42
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Hi ananya022001,

Yes, there is a clean equation route, and it does not get harder as the numbers grow. One shorthand to keep this compact: I will write n ÷ 7 → R4 for "n leaves remainder 4 when divided by 7." For each method below, here is the principle it runs on, then the execution.

Method 1: exploit 63 = 7 x 9

The principle in play:

- A remainder statement is an equation. "n ÷ 7 → R4" says precisely that n = 7a + 4 for some whole number a. Nothing is lost in that translation, so any remainder the question gives can enter the algebra directly.
- A multiple of the divisor carries no remainder. In an expression like 21a + 20, the term 21a is a multiple of 7 and so contributes 0 to the remainder on division by 7. This is what lets us discard whole terms and read a remainder off an expression without ever knowing a.
- Coprime divisors pin down their product. Here 63 = 7 x 9, and 7 and 9 share no common factor. For such a pair, a number's remainders against 7 and against 9 together permit exactly one remainder against 63. That is why the two facts the question hands us are sufficient, and why n itself never has to be found.

Execution:

- n = 7a + 4 gives 3n + 8 = 3(7a + 4) + 8 = 21a + 20. Drop 21a, and 20 = 14 + 6, so (3n + 8) ÷ 7 → R6.
- n = 9b + 5 gives 3n + 8 = 3(9b + 5) + 8 = 27b + 23. Drop 27b, and 23 = 18 + 5, so (3n + 8) ÷ 9 → R5.

So we need the choice that is R6 against 7 and R5 against 9:

- A. 17 = 14 + 3 → R3 against 7. Out.
- B. 20 = 14 + 6 → R6 (good), but 20 = 18 + 2 → R2 against 9. Out.
- C. 26 = 21 + 5 → R5 against 7. Out.
- D. 41 = 35 + 6 → R6, and 41 = 36 + 5 → R5. Passes both.
- E. 62 = 56 + 6 → R6 (good), but 62 = 54 + 8 → R8 against 9. Out.

Therefore only D survives, with no trial and error.

Method 2: solve for n

The principle in play:

- A general form absorbs one condition. Writing n = 9k + 5 makes the ÷9 condition automatic for every whole k. Two conditions on n therefore become one condition on k, trading a large unknown for a small one.
- Splitting a coefficient shrinks the condition. Since 9k = 7k + 2k and 7k is a multiple of 7, testing 9k + 5 against 7 is the same as testing 2k + 5 against 7. That is the same drop principle as above, used here to cut the numbers down rather than to read a remainder off.
- Solutions repeat, so the smallest one suffices. Values of n meeting both conditions recur every 63, the smallest number both 7 and 9 divide. Adding 63 to n adds 3(63) to 3n + 8, which leaves its remainder on division by 63 untouched. So the least valid n gives the same answer as any other.

Execution:

9k + 5 = 7k + (2k + 5), so the ÷7 condition collapses to (2k + 5) ÷ 7 → R4, i.e. 2k + 5 = 7t + 4, so 2k = 7t - 1.

The left side is even, so t must make 7t - 1 even. t = 1 gives 2k = 6, so k = 3.

Back-substituting: n = 9(3) + 5 = 32, so 3n + 8 = 104 = 63(1) + 41, remainder 41, which is D again.

Why this scales

Both methods rest on the same idea: a remainder condition is an equation, and any multiple of the divisor inside it can be thrown away. Each condition therefore collapses into one small statement about one unknown, however large the starting numbers are. Nothing in the method grows with the size of n, which is exactly the guarantee trial and error cannot give you.

Answer: D

ananya022001
Is there any way to solve this using equations? Trial and error of all possible numbers may not work if the number is much bigger right?
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