Question

A five digit number N has all digits different and contains digits 1,3,4,5, and 6 only.
If N is the smallest possible number such that it is divisible by 11, then what is the hundreds place digit of N.​

Answer 1

Question :- A five digit number N has all digits different and contains digits 1,3,4,5, and 6 only. If N is the smallest possible number such that it is divisible by 11, then what is the hundreds place digit of N. ?

Solution :-

we know that, A number is divisible by 11, when :-

• Difference between alternating digits is divisible by 11 or equal to 0 .

so,

• One's digit + Hundred's digit + Ten Thousand's digit - Ten's digit - Thousandth's digit should be divisible by 11 or equal to 0 .

now, as we can see,

→ 1 + 3 + 4 + 5 + 6 = 19 = sum of all digits.

conclusion :-

• Difference between odd digits - Difference between even digits = 11 (0 and 22 not possible.)
• sum of odd digits must be = 15 (4,5 and 6) .
• sum of even digits must be = 4 (1 and 3) .
• Number must be start with 4 . (odd place) .

therefore, total possible numbers formed :-

• 41536 ÷ 11 = Remainder 0 .
• 43516 ÷ 11 = Remainder 0 .
• 41635 ÷ 11 = Remainder 0 .
• 43615 ÷ 11 = Remainder 0 .

hence,

→ Smallest Possible Number = N = 41536 .

∴ The hundreds place digit of N is 5 .

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