# math

i posted this question yesterday. i really want to know how to do it.

John bought n objects, each costing n cents. The change he received from his \$10 bill was less than \$1 and consisted of only two denominations of coins. What was the minimum number of coins that John could have received?

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asked by Diana
1. Lets work in cents.

If he offered 1,000 cents and received less than 100 cents in change then the cost must have been 901 to 999. Can't be 900 as LESS than 100 cents in change and can't be 1000 cents as he received change.

If it is n objects at n cents each then we are looking for a square(s) between 901 and 999.

There is only one square between 901 and 999 which is 31x31=961

Thus the change is 39 cents.

So what was the minimum number of coins that John could have received?

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posted by Dr Russ
2. 1000-31^2=39 (n=31)

39=1*25+14*1 15 coins
39=3*10+9*1 12 coins
39=7*5+4*1 11 coins

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posted by Mgraph
3. Is the minimum amount 6 coins? to equal the 39 cents. 1 quarter, 1 dime and 4 pennies. 25+10+4=39

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posted by Diana
4. 1*25+1*10+4*1=39 in six coins and three denominations.

That's what most cashiers would give you as change, using the "greedy algorithm", i.e. use the biggest possible "denomination" at all times before switching to the lower one.

However, the question requires the change to be in two denominations. Mgraph has enumerated all possible cases for two denominations: make you choice!
Note: cases that do not involve pennies are not possible.

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