Question 536131


im a 3-digit number divisible by 3.my tens digit is 3 times greatas my hundreds digit, and the sum of digits is 15. if you reverse my digits,im divisible by 6,as well as by 3.
what number am i?


Let the hundreds, tens, and units digits be H, T, and U, respectively


Sum of digits = 15, so H + T + U = 15 ----- H + 3H + U = 15 ----- 4H + U = 15


Since T = 3H, then possible H, or hundreds digits are: 1, 3, or 2


If H = 1
4H + U = 15 = 4(1) + U = 15 ----- 4 + U = 15 ----- U = 15 – 4, or 11 (IMPOSSIBLE)


If H = 3
4H + U = 15 = 4(3) + U = 15 ----- 12 + U = 15 ----- U = 15 – 12, or 3 (POSSIBLE)


If H = 3, and U = 3, then we have: H + 3H (T) + U = 15 ---- 3 + 3(3) + 7 = 15 (FALSE)


If H = 2
4H + U = 15 = 4(2) + U = 15 ----- 8 + U = 15 ----- U = 15 – 8, or 7 (POSSIBLE)

If H = 2, and U = 7, then we have: H + 3H (T) + U = 15 ---- 2 + 3(2) + 7 = 15 (TRUE) 
This makes the number: {{{highlight_green(267)}}}

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Check
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This 3-digit number, 267, is indeed divisible by 3 as its sum 15 (2 + 6 + 7) is divisible by 3.
Also, if this 3-digit number is reversed (762), it is indeed divisible by 6 as well as 3.


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