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July 30, 2015

July 30, 2015

Posted by **carrie** on Sunday, February 10, 2013 at 9:06pm.

- tudor -
**bobpursley**, Sunday, February 10, 2013 at 9:25pmI assume you mean the sum of digits.

let the number be xyz

where x is 1

(1+y+z)= 1yz/19

Look at the right side. That has to be a whole number, so 1yz is a multiple of 19

I can range from 6 to 9

which means y+z can range from 5 to 8

try 5=y+z or y=5-z which means if

z; y;

2;3

3,2

4,1

1,4

5,0

0,5

but 1yz=19*6 or 1yz/6=19, which means 1yz is divisible by six, so it ends in 6, 2, 8, 4

if =2, y=4 142 /19=not six

if z=4, y=2 124/19=not six

so yz now runs from 7 to 8

If 1+y+z= then combinations

y+z=6 or 7

assume y+z=6

y, z

0,6

6,0

1,5

5,1

4,2

2,4

Now try these combinations:

1+0+6=7, 7*19=133 nope

1+1+5=7, 133 does not match

none of the y,z above will give 133

Now try y+z=7

6,1 1+6+1=8, 8*19=152 and I see immediately that y=5, z=2

152 is the number

sum digits: 8

the number is 19 times the sum