Mazie counted her dimes. When she put them in groups of 4, she had 2 left over. When she put them in groups of 5, she had one left over. What is the smallest number of dimes she could have, if she has more than 10

14 dimes

how did you come up with this answer

It looks like Rinzin made 3 piles of 4 = 12 and with 2 over - 14. If you only made 2 piles of 4 = 8 with 2 over = 10

BUT the "rules" say more than 10.

Now, if you take 2 piles to 5 (10) with 1 left over = 11.

Sra

IF, however it is the same amount of dimes and you can do BOTH things:

11 piles of 4 = 44 + 2 = 46 and then
9 piles of 5 = 45 + 1 = 46

The answer must be 46.

Sra

46

26

To find the smallest number of dimes Mazie could have, let's approach this problem step by step:

1. Start with a number greater than 10. Let's try 11.

2. Divide 11 by 4 to see if it leaves a remainder of 2 (when put in groups of 4):
11 ÷ 4 = 2 remainder 3

Since the remainder is 3, this is not the correct number of dimes.

3. Now, let's try 12 dimes.

4. Divide 12 by 4:
12 ÷ 4 = 3 remainder 0

Since the remainder is 0, 12 is a possible number of dimes.

5. Next, check if 12 dimes leave a remainder of 1 when put in groups of 5:
12 ÷ 5 = 2 remainder 2

Since the remainder is 2, this is not the correct number of dimes.

6. Continue this process, increasing the number of dimes by 1 each time and repeating steps 4 and 5, until we find a number that satisfies both conditions.

Now, let's continue increasing the number of dimes:

7. Trying 13 dimes:

13 ÷ 4 = 3 remainder 1

13 ÷ 5 = 2 remainder 3

Since 13 does not satisfy both conditions, we move on.

8. Trying 14 dimes:

14 ÷ 4 = 3 remainder 2

14 ÷ 5 = 2 remainder 4

Since 14 does not satisfy both conditions, we move on.

9. Trying 15 dimes:

15 ÷ 4 = 3 remainder 3

15 ÷ 5 = 3 remainder 0

Since 15 satisfies both conditions (leaves a remainder of 2 when divided by 4 and a remainder of 1 when divided by 5), this is the smallest number of dimes Mazie could have.

Therefore, the smallest number of dimes Mazie could have is 15.