Value of U is? P, I and U are digits of two digit numbers
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A two digit number is equal to the sum of its product of digits and sum of digits. How many such two digit numbers are possible?
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Edward Sherry
Answered July 31, 2018
Denote the number by 10T+U, where T is the digit in the tens place and U is the digit in the one's place. The product of its digits is TU; the sum of its digits is T+U. We're given 10T + U = TU + T + U. Subtract U from both sides; we then have 10T = TU + T = T (U +1). Divide both sides through by T (which is legitimate because if T were 0, the number would only be a one-digit number); we then have 10 = U +1, or U = 9. Possible candidates are then 19, 29, 39, etc. Checking these, we see that 19 works: TU = 1*9 = 9; T+U = 2+9 = 10; 19 = 1*9 + 1 + 9 = 9+1+9 = 19 as required. 29 works; TU = 2*9 = 18, T+U = 2+9 = 11; 29 = 18+11 as required. For 39, TU = 3*9 = 27; T+U = 12; 39 = 27 +12 as required. For 49, TU = 4*9 = 36; T+U = 4+9 = 13; 49 = 36 +13 as required. For 59, TU = 5*9 = 45; T+U= 5+9 = 14; 59 = 45+14 as requires. For 69, TU = 6*9 = 54; T+U = 15; 54 +15 = 69 as required. For 79, TU = 7*9 = 63; T+U = 7+9 = 16; 79 = 63 + 16 as required. For 89, TU = 8*9 = 72; T+U = 8+9 = 17; 89 = 72 + 17 as required. For 99, T*U = 9*9 = 81; T+U = 9+9 = 18; 99 = 81 +18 as required.
So all of 19, 29, 39, 49, 59, 69, 79, 89 and 99 qualify.