If 436P5 is divisible by 3, then the largest value of P is:

Correct — C. A number is divisible by 3 exactly when its digit sum is.

Digit sum of 436P5 = 4 + 3 + 6 + P + 5 = 18 + P
18 is already a multiple of 3, so the condition reduces to 3 divides P
Single digits that qualify: 0, 3, 6, 9

The question asks for the largest, which is 9, and 43695 ⁄ 3 = 14565 exactly → option (c).

Why the others are wrong

• (a)
  8 — 8 gives a digit sum of 18 + 8 = 26, and 26 is not a multiple of 3 — 43685 leaves a remainder of 2 on division by 3.
• (b)
  12 — P is a single digit inside the five-figure numeral 436P5, so it cannot be 12. Rule this one out before doing any arithmetic.
• (d)
  6 — 6 does satisfy the rule, since 18 + 6 = 24 is a multiple of 3 — but the question asks for the largest qualifying digit, and 9 works too.

Divisibility by 3 depends only on the sum of the digits, because 10, 100, 1000 and so on all leave a remainder of 1 when divided by 3. Place value therefore drops out entirely.

Here the four fixed digits already sum to 18, itself a multiple of 3. That makes the test unusually clean: the whole condition collapses to whether P is a multiple of 3.

The second half of the question is a reading test. Four digits qualify, and the word largest picks 9; a differently worded stem asking for the least would take the same list and pick 0.

Key facts

• A number is divisible by 3 if and only if the sum of its digits is divisible by 3.
• In 436P5 the fixed digits 4, 3, 6 and 5 sum to 18, a multiple of 3.
• The digits that make 18 + P a multiple of 3 are 0, 3, 6 and 9.
• The same digit-sum test with 9 in place of 3 gives the divisibility rule for 9.

Common traps

• Applying the divisibility rule for 9, which would allow only P = 0 or P = 9.
• Answering the smallest qualifying digit, 0, when the stem asks for the largest.
• Choosing 12 without noticing that P has to be a single digit.

No directly related past PYQ was found.