# Week 1

Ex 50. How many five-digit multiples of 11 are there, if the five digits are 3, 4, 5, 6, 7 in some order?

If the five-digit number abcde (with a + c + e + b + d = 25) is a multiple of 11, then (a + c + e) – (b + d) is a multiple of 11. So, it must be 0, 11 or –11 in the present case.

If (a + c + e) – (b + d) = 0, we get (a + c + e) + (b + d) = 2(b + d), which is impossible because the sum on the left = 25, an odd number and the right side is even.

If (a + c + e) – (b + d) = 11, we get 14 = (a + c + e + b + d) – 11 = 2(b + d). Hence, b + d = 7. The only possible values of b, d are 3, 4 and that of a, c, e are 5, 6, 7 in some order. There are 2 × 3! = 12 such numbers.

If (a + c + e) – (b + d) = –11, we get 36 = (a + c + e + b + d) + 11 = 2(b + d). Hence, b + d = 18, which is not possible.

In the adjacent figure, can the numbers 1, 2, 3, 4, · · · , 18 be placed, one on each line segment, such that the sum of the numbers on the three line segments meeting at each point is divisible by 3?

We group the numbers 1 to 18 in to 3 groups: those leaving remainder 0 when divided by 3; those leaving remainder 1; and those leaving remainder 2. Thus the groups are:
{3, 6, 9, 12, 15, 18}, {1, 4, 7, 10, 13, 16}, {2, 5, 8, 11, 14, 17}
Now we put the numbers in such a way that each of the three line segments converging to a vertex gets one number from each set. For example, here is one such arrangement: