Solving a Divisibility Problem: Finding the Number of Blocks Jeff Had

Jeff had fewer than 100 blocks. When he arranged them into 4 equal rows, he had 3 blocks left over. If he arranged them into 9 equal rows, he had none left over. How many blocks did Jeff have?

Understanding the Problem

The problem requires finding a number N that satisfies the following conditions:

N is less than 100. N leaves a remainder of 3 when divided by 4. This can be mathematically expressed as N ≡ 3 (mod 4). N is divisible by 9. This can be expressed as N ≡ 0 (mod 9).

Step-by-Step Solution

Let's break down the problem into steps to understand how we can find N.

Step 1: Expressing the Conditions Mathematically

From the second condition, we can write:

[ N 4k - 3 ]where k is an integer.

From the third condition, we know:

[ N 9m ]where m is an integer.

Step 2: Equating the Two Expressions for N

Setting the two expressions for N equal gives us:

[ 4k - 3 9m ]Rearranging this, we obtain:[ 4k 9m 3 ]This simplifies to:[ 9m 3 equiv 0 pmod{4} ]

Step 3: Simplifying the Modular Condition

To simplify 9m 3 ≡ 0 (mod 4):

Since 9 ≡ 1 (mod 4), we can replace 9m with m:[ m 3 equiv 0 pmod{4} ]This simplifies to:[ m equiv 1 pmod{4} ]

Thus, we can express m as:[ m 4j - 3 ]where j is an integer.

Step 4: Substituting Back to Find N

Substituting m back into the equation for N:[ N 9m 9(4j - 3) 36j - 27 ]

Step 5: Finding Values of N Under 100

We need to find values of N such that:[ 36j - 27 leq 100 ]Solving for j:[ 36j leq 127 ][ j leq frac{127}{36} approx 3.528 ]

Thus, j can be 0, 1, 2, or 3.

Step 6: Calculating N for Valid j Values

For j 0:[ N 36 times 0 - 27 -27 ] (not valid)

For j 1:[ N 36 times 1 - 27 9 ] (not valid)

For j 2:[ N 36 times 2 - 27 45 ] (not valid)

For j 3:[ N 36 times 3 - 27 81 ] (valid)

Step 7: Verifying the Conditions

Let's verify the conditions for N 81:

When N 81 is divided by 4, the remainder is 1 (81 ÷ 4 20 R1, not 3, so 81 is not a solution). When N 81 is divided by 9, the remainder is 0 (81 ÷ 9 9 R0, which is 0). So 81 meets the second condition.

We need to recheck the valid values. The correct values are:

N 27 N 63 N 99

Step 8: Final Validation and Conclusion

Let's reverify the conditions:

For N 27: 27 ÷ 4 6 R3 (satisfies N ≡ 3 (mod 4)) 27 ÷ 9 3 R0 (satisfies N ≡ 0 (mod 9)) For N 63: 63 ÷ 4 15 R3 (satisfies N ≡ 3 (mod 4)) 63 ÷ 9 7 R0 (satisfies N ≡ 0 (mod 9)) For N 99: 99 ÷ 4 24 R3 (satisfies N ≡ 3 (mod 4)) 99 ÷ 9 11 R0 (satisfies N ≡ 0 (mod 9))

Therefore, the possible values of blocks Jeff could have are 27, 63, or 99. Given that Jeff had fewer than 100 blocks, all three are valid solutions. However, the largest valid solution is:

[ boxed{99} ]

Thus, Jeff had 99 blocks.