Which of the following interchanges of numbers (not digits) would make the given equation correct?

Correct — D. The equation in the figure reads 2^(57 − 33 ÷ 11 × 13 − 2) + 3^(64 − 56 ÷ 8 × 4 − 44) = 35, and 35 splits as 8 + 27 = 2³ + 3³ — so the job is to make both exponents come out as 3.

Swap 33 and 44.
First exponent: 44 ÷ 11 = 4, then 4 × 13 = 52, then 57 − 52 − 2 = 3.
Second exponent: 56 ÷ 8 = 7, then 7 × 4 = 28, then 64 − 28 − 33 = 3.

2³ + 3³ = 8 + 27 = 35 — option (d).

Why the others are wrong

• (a)
  2 and 4 — Swapping the standalone 2 and 4 leaves the exponents at 14 and 6, so the left side is 2¹⁴ + 3⁶ = 16384 + 729 = 17113.
• (b)
  56 and 57 — Swapping 56 and 57 puts 57 ÷ 8 inside the second exponent, and 57 ÷ 8 = 7.125 is not a whole number, so the second term never lands on a power of 3.
• (c)
  57 and 64 — Swapping 57 and 64 lifts the first exponent to 64 − 39 − 2 = 23, and 2²³ alone is over eight million.

The rider numbers, not digits, is doing real work: 33 and 44 move as whole tokens, so you never break them into single 3s and 4s.

Inside each exponent the ordinary order of operations applies — division and multiplication left to right, then the subtractions. So 57 − 44 ÷ 11 × 13 − 2 is 57 − 52 − 2, not (57 − 44) ÷ 11.

The quick route is backwards. Read the target 35, split it into a power of 2 plus a power of 3, and you know what each exponent has to be before you test a single option.

Key facts

• 35 = 8 + 27 = 2³ + 3³, which is the split the swap has to produce.
• Numbers, not digits means the swap moves whole numbers such as 33 and 44.
• Division and multiplication are settled left to right before the subtractions inside each exponent.

Common traps

• Swapping the digits 3 and 4 instead of the numbers 33 and 44.
• Doing the subtraction before the division inside an exponent.
• Testing all four options from the top instead of first working out what the exponents must be.

No directly related past PYQ was found.