Question 106472: find i six digit number in wich the first digit is one less than the second, the secound one less than the third, the fourth digit is one less than the fifth, and the fifth one is one less than the last. the second digit equals the sum of the last three digits and the third digit equals the sum of the secound and fourth digit.
Found 2 solutions by solver91311, edjones:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website! Call the digits d1, d2, d3, d4, d5, and d6
The relationships described are:
d1 = d2 - 1
d2 = d3 - 1
d4 = d5 - 1
d5 = d6 - 1
d2 = d4 + d5 + d6
d3 = d2 + d4
Giving us 6 linear equations in 6 variables.
Rearrange each equation so that you have an expression set equal to zero
1) d1 - d2 + 1 = 0
2) d2 - d3 + 1 = 0
3) d2 - d4 - d5 - d6 = 0
4) -d2 + d3 -d4 = 0
5) d4 - d5 + 1 = 0
6) d5 - d6 + 1 = 0
Add equation 4) to equation 2), resulting in
0d2 + 0d3 - d4 + 1 = 0, which reduces to
d4 = 1
Substituting d4 in equation 5), d5 = 2
Substituting d5 in equation 6), d6 = 3
Substituting d4, d5, and d6 in equation 3), d2 = 6
Substituting d2 in equation 1), d1 = 5
Substituting d2 in equation 2), d3 = 7
Hence the number is 567123.
Check,
5 is one less than 6.
6 is one less than 7.
1 is one less than 2.
2 is one less than 3.
6 is equal to 1 + 2 + 3
7 is equal to 6 + 1
Answer by edjones(8007)  (Show Source):
You can put this solution on YOUR website! a=b-1
b=c-1
d=e-1
e=f-1
b=d+e+f
.
d+e+f<10 So must be 1+2+3 or 2+3+4
b= 6 because if it were 9 then c would be 10
So, d+e+f=6
b=6
d=1
e=2
f=3
a=5
c=7
.
The number is: 567123
We didn't need the final clue.
Ed
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