Numbers and Divisibility

Using the property $a^n - b^n$ is divisible by $(a - b)$, $5^6 - 1$ is divisible by $(5^2)^3 - 1^3$, which is divisible by $5^2 - 1 = 24$, or $(5^3)^2 - 1^2$, divisible by $5^3 - 1 = 124$. Since $124 = 31 \times 4$, the expression is divisible by 31.

Sum of 1 to 100 = 100×101/2 = 5050. 5050 is even, so divisible by 2. 5050/4 = 1262.5 (not integer). 5050/8 = 631.25 (not integer). Therefore 5050 is divisible by 2 only, not by 4 or 8.

The smallest number divisible by 1 through 10 is the LCM of these numbers. Prime factorization: 1, 2, 3, 4=2², 5, 6=2×3, 7, 8=2³, 9=3², 10=2×5. The LCM takes the highest power of each prime: 2³ × 3² × 5 × 7 = 8 × 9 × 5 × 7 = 2520. Options A, B, and C are all missing factors.

The smallest number divisible by all integers from 1 to 10 is the Least Common Multiple (LCM) of those numbers. LCM(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) = 2^3 * 3^2 * 5 * 7 = 8 * 9 * 5 * 7 = 2520.

A number ending in 1 in base 5 and 7 must be of the form 35k + 1. It must also be divisible by 3. Checking 35k+1 for divisibility by 3: k=1 (36), k=4 (141), k=7 (246), etc. In the range 100-999, the values are 141, 246, 351, 456, 561, 666, 771, 876, 981, totaling 9 elements.

Find three-digit numbers where: divisible by 3, ends in 1 in base 5 (so ≡ 1 mod 5), ends in 1 in base 7 (so ≡ 1 mod 7). By CRT, x ≡ 1 (mod 35). Three-digit candidates: 106, 141, 176, 211, 246, 281, 316, 351, 386, 421, 456, 491, 526, 561, 596, 631, 666, 701, 736, 771, 806, 841, 876, 911, 946, 981. Only those divisible by 3 are valid: 141, 81, 246, 351, 561, 666, 841, 946 - totaling 9 numbers.

720 = 2^4 × 3^2. For 720^n to divide 77!, we need at least 4n factors of 2 and 2n factors of 3 and n factors of 5. Count of 5s in 77! is floor(77/5) + floor(77/25) = 15 + 3 = 18. Count of 3s is floor(77/3) + floor(77/9) + floor(77/27) + floor(77/81) = 25 + 8 + 2 + 0 = 35. We need 2n ≤ 35 and n ≤ 18, giving n ≤ 17 (since 2×18 = 36 > 35).

To find the maximum n such that 77! is divisible by 720^n, we factor 720 = 2^4 × 3^2 × 5. In 77!, the exponent of 2 is 73, of 3 is 35, and of 5 is 18. For 720^n = 2^(4n) × 3^(2n) × 5^n, we need 4n ≤ 73, 2n ≤ 35, and n ≤ 18. The limiting factor is 3, giving n ≤ 17 (since 35/2 = 17.5). Thus, maximum n = 17.

The numbers form an arithmetic progression starting at 102 and ending at 396. Using $a=102$, $l=396$, and $d=6$, the number of terms $n$ is $\frac{396-102}{6} + 1 = 50$. The sum is $S_n = \frac{n}{2}(a + l) = \frac{50}{2}(102 + 396) = 25 \times 498 = 12450$. Other options result from calculation errors.

If k is divisible by 2, 3, and 15, it must be a multiple of their LCM, which is 30. For (k + x) to be divisible by 30, x must also be a multiple of 30. Only k+30 satisfies this condition.

A number is divisible by 3 if the sum of its digits is divisible by 3. For the sum to be divisible by 3, the total digit sum including the unknowns (*) must be a multiple of 3. Testing the options, 5 is the maximum value that satisfies the condition.

Factor out 3^71: 3^71(1 + 3 + 9) = 3^71 × 13. Since 13 divides the expression, and 39 = 3 × 13, the expression is divisible by 39. Check: 34, 26, and 19 are not factors of 13 or 3.

The LCM of 7, 10, 15, 21, and 28 is 420. To find the greatest five-digit number, divide 99999 by 420 (remainder 39). 99999 - 39 = 99960. This number is exactly divisible by all the given factors, unlike the other options.

10^5 is 100,000. 100,000 - 7 = 99,993. A number is divisible by 3 if the sum of its digits is divisible by 3. 9+9+9+9+3 = 39, which is divisible by 3. It is odd, so not divisible by 2 or 8.

A number is divisible by 4 if its last two digits ($ab$) form a number divisible by 4. Given $a + b = 11$ and $a < b$, possible pairs $(a,b)$ are $(2,9), (3,8), (4,7), (5,6)$. Among these, only $56$ is divisible by 4. Thus $a=5, b=6$ is the correct solution.