Find a 4-digit odd number using digits 1, 2, 4 and 5 once each such that interchanging the first and last digits gives a number divisible by 4.
One example: 2415
Goal: Make a 4-digit odd number with digits 1, 2, 4, 5 used once each. When we swap the first and last digits, the new number must be divisible by 4.
Divisibility by 4 rule:
A number is divisible by 4 if its last two digits form a multiple of 4.
In symbols:
\( \, ext{if last two digits} = ab,\,\text{ then } 10a+b \text{ is a multiple of } 4. \)
Set up the digits:
Let the original number be \(\overline{A\,B\,C\,D}\).
It must be odd, so \(D\) is odd.
From digits {1, 2, 4, 5}, the odd digits are 1 and 5.
After swapping first and last digits:
New number becomes \(\overline{D\,B\,C\,A}\).
For divisibility by 4, its last two digits \(\overline{C\,A}\) must form a multiple of 4.
Find all possible \(\overline{C\,A}\) from {1, 2, 4, 5} that are multiples of 4:
Try 2-digit combos (no repetition):
So allowed pairs are: \(\overline{C\,A} \in \{12, 24, 52\}.\)
Build valid 4-digit odd numbers for each case:
Case 1: \(\overline{C\,A} = 12\) ⇒ \(C=1,\,A=2\).
Remaining digits: 4 and 5. Since \(D\) must be odd, take \(D=5\). Then \(B=4\).
Number: \(\overline{A\,B\,C\,D} = 2415\).
Swap first and last → \(\overline{D\,B\,C\,A} = 5412\).
Last two digits are \(12\) → divisible by 4 ✔️
Case 2: \(\overline{C\,A} = 24\) ⇒ \(C=2,\,A=4\).
Remaining digits: 1 and 5. \(D\) odd ⇒ \(D=1\) or \(D=5\).
Case 3: \(\overline{C\,A} = 52\) ⇒ \(C=5,\,A=2\).
Remaining digits: 1 and 4. \(D\) odd ⇒ \(D=1\). Then \(B=4\).
Number: \(2451\). Swap → \(1452\), last two \(52\) ✔️
Conclusion:
Examples that satisfy the conditions:
\(2415,\; 4125,\; 4521,\; 2451\).
The sample answer \(2415\) is correct because swapping gives \(5412\) with last two digits \(12\), a multiple of 4.