Step 1

Let x be the number of cards in the box.

Also, the value of x is less than 200.

When x cards are sorted into equal groups of 7, 6 cards were left over.

This implies that x when divided by 7 leaves remainder 6.

That is, x-6 is a multiple of 7.

First, list all multiples of 7 less than 200.

They are 0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 133, 140, 147, 154, 161, 168, 175, 182, 189, 196.

The above list gives the possible values of x-6.

To get the list of possible values of x, add 6 to each number in the above list.

The list of possible values of x is 6, 13, 20, 27, 34, 41, 48, 55, 62, 69, 76, 83, 90, 97, 104, 111, 118, 125, 132, 139, 146, 153, 160, 167, 174, 181, 188, 195, 202.

Step 2

Further, when x cards are sorted into equal groups of 5, 4 cards were left over.

This implies that x when divided by 5 leaves remainder 4.

If x leaves remainder 4 when divided by 5, then x will have either 4 or 9 in the units place.

List out the numbers with 4 or 9 in the units place from the list of possible values of x.

The corresponding list of possible values of x is 34, 69, 104, 139, 174.

The last condition is that when x cards are sorted into equal groups of 3, 2 cards were left over.

This implies that x when divided by 3 leaves remainder 2.

The only value of x from the list 34, 69, 104, 139, 174 that leaves remainder 2 when divided by 3 is 104.

It can be noted that \(\displaystyle{104}={3}\times{34}+{2}\).

Thus, the only possible value of x satisfying all the given conditions is 104.

Hence, the number of cards that were in the box is 104.

We can verify that 104 satisfies all the given conditions as shown below.

Clearly, 104 is less than 200.

Since \(\displaystyle{104}={7}\times{14}+{6}\), it leaves remainder 6 when divided by 7.

Since \(\displaystyle{104}={5}\times{20}+{4}\), it leaves remainder 4 when divided by 5.

Since \(\displaystyle{104}={3}\times{34}+{2}\), it leaves remainder 2 when divided by 3.

Let x be the number of cards in the box.

Also, the value of x is less than 200.

When x cards are sorted into equal groups of 7, 6 cards were left over.

This implies that x when divided by 7 leaves remainder 6.

That is, x-6 is a multiple of 7.

First, list all multiples of 7 less than 200.

They are 0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 133, 140, 147, 154, 161, 168, 175, 182, 189, 196.

The above list gives the possible values of x-6.

To get the list of possible values of x, add 6 to each number in the above list.

The list of possible values of x is 6, 13, 20, 27, 34, 41, 48, 55, 62, 69, 76, 83, 90, 97, 104, 111, 118, 125, 132, 139, 146, 153, 160, 167, 174, 181, 188, 195, 202.

Step 2

Further, when x cards are sorted into equal groups of 5, 4 cards were left over.

This implies that x when divided by 5 leaves remainder 4.

If x leaves remainder 4 when divided by 5, then x will have either 4 or 9 in the units place.

List out the numbers with 4 or 9 in the units place from the list of possible values of x.

The corresponding list of possible values of x is 34, 69, 104, 139, 174.

The last condition is that when x cards are sorted into equal groups of 3, 2 cards were left over.

This implies that x when divided by 3 leaves remainder 2.

The only value of x from the list 34, 69, 104, 139, 174 that leaves remainder 2 when divided by 3 is 104.

It can be noted that \(\displaystyle{104}={3}\times{34}+{2}\).

Thus, the only possible value of x satisfying all the given conditions is 104.

Hence, the number of cards that were in the box is 104.

We can verify that 104 satisfies all the given conditions as shown below.

Clearly, 104 is less than 200.

Since \(\displaystyle{104}={7}\times{14}+{6}\), it leaves remainder 6 when divided by 7.

Since \(\displaystyle{104}={5}\times{20}+{4}\), it leaves remainder 4 when divided by 5.

Since \(\displaystyle{104}={3}\times{34}+{2}\), it leaves remainder 2 when divided by 3.