The sum of the digit of a three-digit number is 11. If the digits are reversed, the new number is 46 more than five time the old number. If the hundreds digit plus twice the tens digit is equal to the units digit, then what is the number?
2007-09-17 05:32:14 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Old number is xyz
The sum of the digit of a three-digit number is 11
1) x + y + z = 11
If the digits are reversed, the new number is 46 more than five time the old number.
2) 100z + 10y + x = 5(100x + 10y + z) + 46
100z + 10y + x = 500x + 50y + 5z + 46
- 499x - 40y + 95z = 46
the hundreds digit plus twice the tens digit is equal to the units digit
3) x + 2y = z
x + 2y - z = 0
With some effort, solve this system of three equations in three unknownx
2007-09-17 05:58:03 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Since five times the number plus 46 is still a 3-digit number, the hundreds digit must be 1.
If b is the tens digit, and c the ones digit, then:
1 + b + c = 11, or b+c = 10 (first condition)
and
1 + 2b = c (third condition)
Putting 1+2b in for c in b+c = 10, we get b=3.
Then c =7.
Now we have to check the middle condition, that:
137 * 5 + 46 = 731
2007-09-17 05:46:52 · answer #2 · answered by thomasoa 5 · 2⤊ 0⤋
The number by a quick guess is 137
2007-09-17 05:47:21 · answer #3 · answered by lonelyspirit 5 · 0⤊ 0⤋
a + b + c = 11
100*c + 10*b + a = 5[100*a + 10*b + c] + 46
a + 2*b = c
With those equations, you should be able to find a, b, and c (the individual digits).
2007-09-17 05:38:23 · answer #4 · answered by Mathsorcerer 7 · 0⤊ 0⤋
I believe the answer is 137
2007-09-17 05:48:50 · answer #5 · answered by John Z 1 · 0⤊ 0⤋