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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?
Found 2 solutions by lwsshak3, MathTherapy:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! 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 u=units digit
let t=tens digit
let h=hundreds digit
..
u+t+h=15
t=3h
possible ten and hundred digits:
t=3, h=1
u+3+1=15
u=15-4=11
u≠11
..
t=6, h=2
u+6+2=15
u=15-8=7
u=7
number:762
Check:
762/3=254
762/6=127
ans:
The number is 762
Answer by MathTherapy(10553)  (Show Source):
You can put this solution on YOUR website!
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: 
-------
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.
Send comments and “thank-yous” to “D” at MathMadEzy@aol.com
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