A three-digit password has all different digits, its first digit is 2 more than its second digit, and its three digits add up to 10. Which is correct: 7-2-1, 4-2-4, 1-3-6, or 5-3-2?

Step-by-step solution

1. 1

   List the three conditions as tests. With digits $d_1d_2d_3$:

   $\text{(i) all different},\qquad \text{(ii) } d_1=d_2+2,\qquad \text{(iii) } d_1+d_2+d_3=10.$

   With four candidates offered, checking each against three short tests is faster than solving from scratch.

2. 2

   Apply the cheapest test first — distinctness. Option B is $4$-$2$-$4$, which repeats the digit $4$. Eliminated immediately, without any arithmetic.

3. 3

   Apply the first-digit rule to the rest.

   $\text{A: }7\overset{?}{=}2+2=4\ \text{✗},\qquad \text{C: }1\overset{?}{=}3+2=5\ \text{✗},\qquad \text{D: }5\overset{?}{=}3+2=5\ \text{✓}.$

   Options A and C are eliminated. Note that all three of A, C and D happen to have digit sum $10$, so condition (iii) alone would not have separated them — that is the trap in the design.

4. 4

   Confirm the survivor against all three conditions. For $5$-$3$-$2$: the digits $5,3,2$ are all different ✓; $5=3+2$ ✓; $5+3+2=10$ ✓.

5. 5

   Solve directly and see which option survives. From (ii), $d_2=d_1-2$; substituting into (iii) gives $d_3=12-2d_1$. Requiring $0\le d_3\le 9$ forces $d_1\in\{2,\ldots,6\}$, and distinctness rules out only $d_1=4$ (which gives $d_3=4=d_1$). That leaves four legal passwords — $208$, $316$, $532$ and $640$ — of which only D: 5-3-2 appears among the options.