Question 802580
(1) {{{ p + 1 = n/2 }}}
(2) {{{ n = d + 1 }}}
(3) {{{ q + 3 = p + n + d }}}
(4) {{{ 1*p + 5n + 10d + 25q = 800 }}} ( in cents )
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I can get (4) in terms of just nickels
(1) {{{ p = n/2 - 1 }}}
and
(2) {{{ d = n - 1 }}}
and
(3) {{{ q = p + n + d  - 3 }}}
(3) {{{ q = n/2 - 1 + n + n - 1 - 3 }}}
(3) {{{ q = (5n)/2 - 5 }}}
and
(4) {{{ 1*p + 5n + 10d + 25q = 800 }}}
(4) {{{ n/2 - 1 + 5n + 10*( n-1 ) + 25*( (5n)/2 - 5 ) = 800 }}} 
(4) {{{ n/2 - 1 + 5n + 10n - 10 + ( 125n )/2 - 125 = 800 }}}
Multiply both sides by {{{ 2 }}}
(4) {{{ n - 2 + 10n + 20n - 20 + 125n - 250 = 1600 }}}
(4) {{{ 156n = 1600 + 2 + 20 + 250 }}}
(4) {{{ 156n = 1872 }}}
(4) {{{ n = 12 }}}
and
(2) {{{ d = n - 1 }}}
(2) {{{ d = 12 - 1 }}}
(2) {{{ d = 11 }}}
There are 11 dimes
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check:
(1) {{{ p = n/2 - 1 }}}
(1) {{{ p = 12/2 - 1 }}}
(1) {{{ p = 5 }}}
and
(3) {{{ q = p + n + d  - 3 }}}
(3) {{{ q = 5 + 12 + 11 - 3 }}}
(3) {{{ q = 25 }}}
and
(4) {{{ 1*p + 5n + 10d + 25q = 800 }}}
(4) {{{ 1*5 + 5*12 + 10*11 + 25*25 = 800 }}}
(4) {{{ 5 + 60 + 110 + 625 = 800 }}}
(4) {{{ 800 = 800 }}}
OK