A six-digit positive integer, ABCDEF, is divisible by all the integers

Solution

• ABCDEF is a six-digit number, and is divisible by all the integers from 2 to 6, both inclusive
• All the six digits are distinct and are less than 7

• The number of such numbers

• ABCDEF is divisible by 2 and 5, which implies that the units digit must be 0

o So, F = 0

• We are given that all the digits are less than 7
• Implies, the possible digits are {0, 1, 2, 3, 4, 5, 6}

o And since, F = 0, no other digit can take that value.
o Sum of the remaining 6 digits of the above set = 1 + 2 + 3 + 4 + 5 + 6 = 21

• Now, A, B, C, D and E can take one value each among these 6 digits, such that their sum is a multiple of 3.

o But, since ABCDEF is divisible by 4, EF must be 20 or 40 or 60
o So, E can take one of the digits among {2, 4, 6}

• The two possible cases for sum to be a multiple of 3 are 21 - 3 and 21 – 6

o That is, (1 + 2 + 4 + 5 + 6) and (1 + 2 + 3 + 4 + 5)

• Case-1: {1, 2, 4, 5, 6}

o Since, E must be even, the number of possibilities for E = 3
o The remaining 4-digits can be arranged in 4! ways = 24
o So, the total number of possibilities = 3 * 24 = 72

• Case 2: {1, 2, 3, 4, 5}

o As E is even, the number of possibilities for E = 2
o The remaining 4-digits can be arranged in 4! ways = 24
o So, the total number of possibilities = 2 * 24 = 48