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moneybird Group Title

Grade 10 Cayley #25 The digits of the positive integer n include no 9s, exactly four 8s, exactly three 7s, exactly two 6s, and some other digits. If the sum of the digits of n is 104 and the sum of the digits of 2n is 100, then the number of times that the digit 5 occurs in n is...?

  • 2 years ago
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  1. rlsadiz Group Title
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    for the sake of argument lets assume that all the digits of n from 1-> 9 be separated by 0's. i.e. 1020304050606 We know that there are 4 8's, 3 7's 2 6's \[104 = 8(4) + 7(3) + 6(2) +X_{1}\]\[X_{1} = 39\]Where X1 is the sum of the digits less that 6. Notice that \[8 * 2 = 16\]\[1 + 6 = 7\] \[7 * 2 = 14\]\[1 + 4 = 5\] \[6 * 2 = 12\]\[1 + 2 = 3\] \[5 * 2 = 10\]\[1 + 0 = 1\] Since we put a 0 between the digits the double of the number will not "overflow" to the other digits. using or example earlier. Twice of it s 2040608101212. So we could say, \[100 = 7(4) + 5(3) + 3(2) + X_{2}\]\[X_{2} = 51\] Note that lets say i have a number A wherein all its digits are less that 5. The sum of digits of 2A is equal to the double of the sums of digits of A because\[4 * 2 = 8\]\[3 * 2 = 6\]\[2 * 2 = 4\]\[1 * 2 = 2\] So we could say that \[X_{1} = a + 5b\]\[X_{2} = 2a + b\] Where a is the sum of all the digits less than 5 and b is the number of 5's in n. Solving for b \[39 = a + 5b\]\[51 = 2a + b\] \[3 = b\] So the answer is there are 3 5's in n.

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