Question 571992
Let the 10's digit = {{{ a }}}
Let the 1's digit = {{{ b }}}
The number is {{{ 10a + b }}}
The number reversed is {{{ 10b + a }}}
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given:
(1) {{{ 10a + b = 2*( a + b ) + 9 }}}
(2) {{{ 10b + a = 8*( a + b ) }}}
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(1) {{{ 10a + b = 2a + 2b + 9 }}}
(1) {{{ 10a - 2a + b - 2b = 9 }}}
(1) {{{ 8a - b = 9 }}}
and
(2) {{{ 10b + a = 8*( a + b ) }}}
(2) {{{ 10b + a = 8a + 8b }}}
(2) {{{ 7a = 2b }}}
(2) {{{ a = (2/7)*b }}}
Substitute this into (1)
(1) {{{ 8a - b = 9 }}}
(1) {{{ 8*(2/7)*b - b = 9 }}}
(1) {{{ 16b - 7b = 63 }}}
(1) {{{ 9b = 63 }}}
(1) {{{ b = 7 }}}
and, since
(2) {{{ a = (2/7)*b }}}
(2) {{{ a = (2/7)*7 }}}
(2) {{{ a = 2 }}}
The number is 27
check:
{{{ 27 = 2*( 2 + 7 ) + 9 }}}
{{{ 27 = 18 + 9 }}}
{{{ 27 = 27 }}}
and
{{{ 72 = 8*( 2 + 7 ) }}}
{{{ 72 = 72 }}}
OK