Question 1047240
Since {{{30}}} nickels make {{{30*("$0.05")="$1.50"}}} ,
you will need to use some pennies.
The number of pennies used has to be a multiple of {{{5}}} ,
because, in cents, all the other coins amount to multiples of 5.
 
Could you do it with just {{{5}}} pennies?
That would leave
{{{"$1.00"-"$0.05"="$0.95"}}}
to be made up with the other {{{30-5=25}}} coins.
Could we find 25 coins whose values add up to {{{95}}} cents?
{{{95}}} cents is {{{95/5=19}}} nickels,
and {{{5 pennies + 19 nickels=24 coins}}} is not enough coins.
You cannot do it with only {{{5}}} pennies.
 
Could you do it with exactly {{{10}}} pennies?
That would leave {{{100-10=90}}} cents to be made up
with the other {{{30-10=20}}} coins.
{{{90}}} cents is {{{90/5=18}}} nickels,
and {{{10 pennies + 18 nickels=28 coins}}} is not enough coins.
 
Could you do it with exactly {{{15}}} pennies?
That would leave {{{100-15=85}}} cents to be made up
with the other {{{30-15=15}}} coins.
{{{85}}} cents is {{{85/5=17}}} nickels,
and {{{15 pennies + 17 nickels=32coins}}} is too many coins,
but that tells you
that with {{{15}}} pennies,
you may be able to make up the remaining {{{85}}} cents
with the {{{15}}} extra coins you are allowed to use.
If you exchange {{{2}}} nickels for {{{1}}} dime,
you get the same amount, with {{{2-1=1}}} less coin.
So, if you do that twice, you end with
{{{17-2-2=highlight(13)}}} nickels, and
{{{1+1=highlight(2)}}} dimes,
along with the {{{highlight(15)}}} pennies,
adding up to {{{13+2+15=30}}} coins,
and an amount, in cents, of
{{{15+13*5+2*10=15+65+20=100}}} (which is $1).
 
OTHER OPTIONS:
Could you use {{{20}}} pennies?
You would need to make up {{{100-20=80}}} cents
with the remaining {{{30-20=10}}} coins.
{{{80}}} cents would be {{{80/5=16}}} nickels.
That is too many coins.
You need {{{16-10=6}}} less coins.
If you get {{{6}}} dimes in exchange for {{{12}}} nickels<
you end up with:
{{{16-12=highlight(4)}}} nickels, and
{{{highlight(6)}}} dimes,
along with the {{{highlight(20)}}} pennies,
adding up to {{{4+6+20=30}}} coins,
and an amount, in cents, of
{{{20+4*5+6*10=20+20+60=100}}} (which is $1).

Also with {{{20}}} pennies, and starting from {{{16}}} nickels, you could instead,
get {{{1}}} quarter in exchange for {{{5}}} nickels,
to use {{{5-1=4}}} less coins,
and further reduce the number of coin
by getting {{{2}}} dimes in exchange for {{{4}}} nickels.
That would leave you with
{{{highlight(1)}}} quarter,
{{{highlight(2)}}} dimes, and
{{{16-5-4=highlight(7)}}} nickels,
along with the {{{highlight(20)}}} pennies,
adding up to {{{1+2+7+20=30}}} coins,
and an amount, in cents, of
{{{20+7*5+2*10+25=20+35+20+25=100}}} (which is $1).
 
Could you do it using exactly 25 pennies?
You would need to make up {{{100-25=75}}} cents
with the remaining {{{30-25=5}}} coins.
{{{75}} cents is {{{75/5=15}}} nickels,
and that is {{{15-5=10}}} coins too many.
You already know that trading coins reduces the number
by {{{4}}} each time you get {{{1}}} quarter for {{{5}}} nickels,
and by {{{1}}} each time you get {{{1}}} dime for {{{2}}} nickels,
so getting {{{2}}} quarters for {{{10}}} nickels,
and {{{2}}} dimes for {{{4}}} nickels, you end up with
{{{highlight(2)}}} quarters,
{{{highlight(2)}}} dimes, and
{{{15-10-4=highlight(1)}}} nickel,
along with the {{{highlight(25)}}} pennies,
adding up to {{{2+2+1+25=30}}} coins,
and an amount, in cents, of
{{{2*25+2*10+5+25=50+20+5+25=100}}} (which is $1).