I am a 3-digit number
All my digits are odd
I am less than 600
I am greater than 400
If you add my digits you get 15
My last digit is 7
What am I?
2006-08-25 14:46:19 · 12 answers · asked by jjohnson52388 3 in Science & Mathematics ➔ Mathematics
Since all the digits have to be odd, the first has to be a 5. The second digit has to be 15-7-5=3. Thus, the answer is 537.
2006-08-25 16:26:31 · answer #1 · answered by Kyrix 6 · 0⤊ 0⤋
Since the number is between 400 and 600,
and all the digits are odd, the first digit must
be a 5. Since the last digit is 7 and the sum
of all the digits is 15 the middle digit must be 3.
Answer: 537
2006-08-25 22:16:40 · answer #2 · answered by steiner1745 7 · 0⤊ 0⤋
537
2006-08-29 21:20:06 · answer #3 · answered by pug hugger 2 · 0⤊ 0⤋
537
2006-08-26 04:47:10 · answer #4 · answered by PenguinMoose 3 · 0⤊ 0⤋
537
2006-08-26 00:36:37 · answer #5 · answered by leshdog300 2 · 0⤊ 0⤋
537
2006-08-25 22:13:47 · answer #6 · answered by rakisup 2 · 0⤊ 0⤋
537
2006-08-25 21:59:35 · answer #7 · answered by hulio g 1 · 0⤊ 0⤋
a, b, c are the numbers where c=7 and a,b are odd
a+b+c = 15, ==> a+b = 15-c = 15-7 = 8
possibilities:
a=1, b=7 and a=3,b=5 or change a with b
but 177 is less than 400
the answer is 537
2006-08-25 21:50:58 · answer #8 · answered by ___ 4 · 0⤊ 0⤋
537
2006-08-25 21:50:31 · answer #9 · answered by In Like Flynn 2 · 0⤊ 0⤋
537
2006-08-25 21:50:09 · answer #10 · answered by iuud2noitall 3 · 0⤊ 0⤋