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# Sandy has a collection of nickels and dimes.?

Sandy has a collection of nickels and dimes. The nickels can be put into piles so that the number of nickels in each pile is the same as the number of piles. Also, the dimes can be put into piles so that the number of dimes in e ach pile is the same as the number of piles. Which of the following could not be the total value of Sandy’s coin collection?

a. \$3.40 b. \$5.35 c. \$3.00 d. \$6.05

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Say there are x nickels and y dimes.

If there are x piles of x nickels, that's x^2 nickels altogether.

Similarly, there are y^2 dimes altogether.

The value of the nickels is 5x^2 cents and the value of the dimes is 10y^2 cents.

The total value is therefore T = 5x^2 + 10y^2 cents = 5(x^2 + 2y^2) cents.

Thus, x^2 + 2y^2 = T/5

I simplified it to this form because each answer is divisible by 5.

Now substitute each answer for T, divide it by 5, and we have these possibilities :

(a) x^2 + 2y^2 = 68

(b) x^2 + 2y^2 = 107

(c) x^2 + 2y^2 = 60

(d) x^2 + 2y^2 = 121

Now look at what numbers x and y can be.

We first assume that x and y are not zero and we can immediately see that y < 8.

The concept for this problem is that when you add or multiply numbers together, the

units digit is all that we need. Note that the units digit in each of (a), (b), (c) and (d)

above, are 8, 7, 0 and 1, respectively.

Firstly, in (b) and (c), the answers 107 and 121, are odd numbers.

Because 2y^2 is always even, then it must be that x is odd.

The units digit of x can then only be 1, 3, 5, 7 or 9.

Now square those, but only retain the units digit.

x^2 = 1, 9, 5, 9 or 1.

Thus, x^2 can only be 1, 5 or 9.

The units digit of y can be any of 1, 2, 3, 4, 5, 6 or 7.

Therefore, the units digit of y^2 is 1, 4, 9, 6, 5, 6 or 9.

Simplifying, y^2 ends in either 1, 4, 5, 6 or 9.

Therefore, 2y^2 ends in either 2, 8, 0, 2 or 8.

Thus, 2y^2 can only be 0, 2 or 8.

Now add all combinations of x^2 (1, 5, 9) plus 2y^2 (0, 2, 8).

All combinations yield the end digit as 1, 3, 5, 7, or 9.

107 and 121 do end with one of these digits, so they may be possibilities.

Secondly, the answers of 68 and 60 are both even numbers, so x must be even.

If x ends in 2, 4, 6 or 8, then x^2 ends in 4, 6, 6, or 4. That is, x = 4 or 6.

2y^2 we know already, ends in 0, 2 or 8.

Adding all combinations of these gives : x^2 + 2y^2 ends in either 2, 4, 6 or 8.

Answer (a) = 68 ends in 8, so this is a possibility.

However, (c) = 60 ends in 0, but we've seen that this is impossible, for the set of

numbers given.

Therefore, (c) \$3.00, is the one that could not possibly be the total value.

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