IPMAT Indore 2023 · Quantitative Ability · Question 30

Step 1 — build the cube table modulo 10.

Notice this map is a bijection — every digit appears exactly once in the second row. That is special to cubes (squares are not like this), and it means the unit digit of p is recoverable from the unit digit of {p}^{3}.

Step 2 — identify p. The table shows {d}^{3} ends in 4 only for d=4:

p\equiv 4 (\mod 10)

Step 3 — shift by 3.

p+3\equiv 7 (\mod 10) \Rightarrow{\left( p+3 \right)}^{3}\equiv{7}^{3}=343\equiv 3 (\mod 10)

Step 4 — compare with the options. Strictly, the only achievable unit digit is 3.

The published key marks Option 1 (1, 3), which is the official answer for this question. The digit 1 is what you get from {\left( p-3 \right)}^{3}: p-3\equiv 1 and {1}^{3}\equiv 1. So the intended reading appears to have been {\left( p\pm 3 \right)}^{3}, and Option 1 collects both outcomes.

Answer: Option 1 — 1, 3 (the official key; see the note below)

Teaching note. Work this through with students rather than around it. Taken exactly as printed, {\left( p+3 \right)}^{3} can only end in 3, and of the four options only (a) and (b) contain 3 at all. Option 1 is the official answer and the one that scores, but the “1” in it comes from p-3, not p+3. It makes a good illustration of why the cube-residue table is worth memorising: it shows immediately that p is pinned to a single residue, which is what makes the question’s own options look over-generous.

Shortcut. Memorise the cube table mod 10 — it is a bijection.

Unlike squares (which only ever end in 0, 1, 4, 5, 6, 9), every digit is the cube of exactly one digit. So “the cube ends in 4” pins p to end in 4 — no case work at all.

The two residues to carry: 4+3=7 and {7}^{3} ends in 3; 4-3=1 and {1}^{3} ends in 1. Only options (a) and (b) contain a 3, and (b) adds a 7 that no shift produces — so (a) is the one to mark, in about fifteen seconds.