Question #d0593

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Question #d0593

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Answer 1 · 2017-10-27T13:44:34

The answer is $344 = 7 \times 49 + 1 = 9 \times 38 + 2 = 11 \times 31 + 3$ $344=7xx49+1=9xx38+2=11xx31+3$

Explanation:

We must this by try and error

let's look for $7 a + 1 = 9 b + 2$ $7a+1=9b+2$

$7 \times 0 + 1 = 1 = 9 \times 0 + 1$ $7xx0+1=1=9xx0+1$
$7 \times 1 + 1 = 8 = 9 \times 0 + 8$ $7xx1+1=8=9xx0+8$
$7 \times 2 + 1 = 15 = 9 \times 1 + 6$ $7xx2+1=15=9xx1+6$
$7 \times 3 + 1 = 22 = 9 \times 2 + 4$ $7xx3+1=22=9xx2+4$
$7 \times 4 + 1 = 29 = 9 \times 3 + 2 \to$ $7xx4+1=29=9xx3+2->$ solution

We can determine that solutions will repeat in a cycle of 63 so we can do like this:

$63 + 29 = 92 = 11 \times 8 + 4$ $63+29=92=11xx8+4$
$92 + 63 = 155 = 14 \times 8 + 1$ $92+63=155=14xx8+1$
$155 + 63 = 218 \Rightarrow 11 \times 19 + 9$ $155+63=218=>11xx19+9$
$218 + 63 = 281 = 11 \times 25 + 6$ $218+63=281=11xx25+6$
$281 + 63 = 344 = 9 \times 38 + 2 \to 11 \times 31 + 3 \to$ $281+63=344=9xx38+2->11xx31+3->$ solution

in fact

$344 = 7 \times 49 + 1 = 9 \times 38 + 2 = 11 \times 31 + 3$ $344=7xx49+1=9xx38+2=11xx31+3$

Other solution can be found by adding $(7 \times 9 \times 11) n$ $(7xx9xx11)n$

$344 + 693 = 1037$ $344+693=1037$
$1037 + 693 = 1730$ $1037+693=1730$

$e t c .$ $etc.$

$693 n + 344$ $693n+344$

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Question #d0593

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