2006-10-25 17:58:17 · 9 answers · asked by Bev 3 in Science & Mathematics ➔ Mathematics
256 / 4 = 64
2 + 5 + 6 = 13
292 / 4 = 73
2 + 9 + 2 = 13
328 / 4 = 82
3 + 2 + 8 = 13
364 / 4 = 91
3 + 6 + 4 = 13
Peace!
2006-10-25 18:02:27 · answer #1 · answered by carole 7 · 1⤊ 0⤋
Let ABC be the number.
If it's between 200 and 300, then A = 2 or 3.
If it's divisible by 4, then C = 0, 2, 4, 6 or 8.
All combinations are then given by :
2B0, 2B2, 2B4, 2B6, 2B8, 3B0, 3B2, 3B4, 3B6 and 3B8.
If the sum of the digits is 13, then 2B0 and 3B0 are impossible.
That leaves the following combinations :
292, 274, 256, 238, 382, 364, 346 and 328.
There is a rule that says - if a number is divisible by 4, then the last 2 digits are divisible by 4.
This is not true for 274, 238, 382 and 346.
Thus the number must be either 292, 256, 364 or 328.
2006-10-26 00:25:37 · answer #2 · answered by falzoon 7 · 0⤊ 0⤋
292
2006-10-25 19:27:33 · answer #3 · answered by Anonymous · 0⤊ 0⤋
292
2006-10-25 18:00:37 · answer #4 · answered by suzyQ 3 · 0⤊ 0⤋
328 So now do I get 10 points. I mean is that the number you were thinking of :)
2006-10-25 18:06:41 · answer #5 · answered by lost_soul 4 · 0⤊ 0⤋
292 is the answr dude
2006-10-25 18:16:49 · answer #6 · answered by Jason 1 · 0⤊ 0⤋
you are 256
2006-10-25 20:01:57 · answer #7 · answered by the walking brother 2 · 0⤊ 0⤋
364. There are others but that jumped into my head. Are you sure you included the whole question?
2006-10-25 18:02:42 · answer #8 · answered by Kuji 7 · 0⤊ 0⤋
256, but I suspect that there are multiple other answers.
2006-10-25 18:01:35 · answer #9 · answered by Tim N 5 · 0⤊ 0⤋