Dudeney, Amusements in Mathematics Problem 12 It will be found that Â£66, 6s. 6d equals 15, 918 pence. Now, the four 6's added together make 24, and the figures 15, 918 also add to 24. It is a curious fact that there is only one other sum of money, in pounds, shillings and pence (all similarly repetitions of one figure), of which the digits shall add up the same as the digits of the amount in pence. What is the other sum of money?
We are looking for k = 1, 2, 3, ... , or 9 and n = 1, 2, 3, ... such that the digit sum of k ... k (n ks) and k and k is the same as that of (240 k ... k + 12k + k). But in the case of k = 1, we can also have 11s 1d, 1s 11d or 11s 11d. The difficulty is that it is not clear how to find the digit sum for the number of pence which is 240 k ... k + 12k + k. Let us take k = 1. Taking 1, 2, 3, 4 digits in the number of pounds we get: 253, 2653, 26653, 266653. The pattern is fairly clear - with n digits we get n-1 6s and a digit sum of 10 + 6(n-1) = 6n + 4. The number of digits in the Â£ s d expression is n + 2, so they can obviously never be the same. To prove the pattern we could use induction. Suppose that it is true for n 1s. If we move to (n+1) 1s we increase k ... k by 10 n and hence 240 k ... k by 240.10 n = 2.10 n+2 + 4.10 n+1 . The value for n 1s is 2.10 n+1 + lower powers of 10, so we change the first digit from 2 to 6 and place an additional 2 in front. In other words, we insert a 6 after the initial 2. That establishes the result. Similarly, for k = 2, we get 506, 5306, 53306, ... so the two digit sums are 11 + 3(n-1) = 3n+8 and 2(n+2). Again they can never be equal. For k = 3, we get 759, 7959, 79959, ... , so the two digit sums are 12 + 9n and 3(n+2). Again they cannot be equal. For k = 4, we get 1012, 10612, 106612, ... , so the two digit sums are 4 + 6(n-1) = 6n-2 and 4(n+2). They are equal if n = 5. So we get the solution Â£44,444 4s 4d = 10,666,612d, both with digit sum 28. For k = 5, we get 1265, 13265, 133265, ... , so the two digit sums are 14 + 3(n-1) = 3n+11 and 5(n+2), which has no solutions. For k = 6, we get 1518, 15918, 159918, ... , so the two digit sums are 15 + 9(n-1) = 9n+6 and 6(n+2), which has the solution n = 2, giving the solution in the question. For k = 7, we get 1771, 18571, 186571, 1866571, 18666571, ... , so the digit sums are 22 + 6(n-2) = 6n+10 (which also works for n = 1) and 7(n+2), so there are no solutions. For k = 8, we get 2024, 21224, 213224, 2133224, 21333224, ... , so the digit sums are 11 + 3(n-2) = 3n+5 (which also works for n =1) and 8(n+2), so there are no solutions. For k = 9, we get 2277, 23877, 239877, 2399877, ... , so the digit sums are 27 + 9(n-2) = 9n+9 (which also works for n = 1) and 9(n+2), so there are no solutions. Finally, we need to check the other cases for k = 1. For 1s 11d, we increase both digit sums by 1, so there are still no solutions. For 11s 1d, we increase the pence digit sum by 3 to 6n+7 and the Â£ s d digit sum by 1 to n+3, so there are still no solutions. For 11s 11d, we increase the pence digit sum by 4 to 6n+8 and the Â£ s d digit sum by 2 to n+4, so there are still no solutions. Summarising, there is only one additional solution, namely Â£44,444 4s 4d = 10,666,612d. Dudeney 1-25 (C) John Scholes jscholes@kalva.demon.co.uk 25 Oct 2002