Divisibility Rules(LEVEL - 2)
Q1–10 of 40A 6-digit number is formed by repeating a 3-digit natural number (e.g., 256256). What is the largest number that always divides such a number?
Hint/Explanation: Split the number by place value: XYZXYZ = XYZ000 + XYZ = XYZ(1000+1) = XYZ x 1001. Since XYZ can be anything, 1001 is the fixed factor that always divides it.
Find the remainder when the 100-digit number 7777...77 is divided by 101.
Hint/Explanation: Just as 11 works on alternate single digits (10^1+1), 101 works on alternate 2-digit blocks (10^2+1). Breaking 100 sevens into 50 blocks of 77 gives an alternating sum of +77-77+77-77... = 0.
A 96-digit number N is formed by writing the digit 7 ninety-six times. Another number M is formed by writing 3 ninety-six times. What is the remainder when N x M is divided by 1001?
Hint/Explanation: 1001 = 7 x 11 x 13. A repunit block with a digit count that is a multiple of 6 (like 96 = 6x16) is completely divisible by 1001. Since N = 0 (mod 1001), the product N x M = 0.
What is the remainder when N = 103 + 106 + 109 + ... + 1099 is divided by 999?
Hint/Explanation: 103 = 1000 = 1 (mod 999). Every term of the form 103k = (103)^k = 1k = 1 (mod 999). Summing 33 such terms gives a remainder of 33 x 1 = 33.
If the 8-digit number 34A56B23 is divisible by 99, find the value of A+B.
Hint/Explanation: Break 99 into co-primes 9 and 11. Divisibility by 9 forces sum of digits 23+A+B to give A+B = 4 or 13. Checking the alternating digit difference for 11 forces A-B = 1, which only works with A+B = 13 (A=7, B=6).
How many 4-digit numbers of the form ABBA are divisible by 101?
Hint/Explanation: Expand algebraically: ABBA = 1001A + 110B = -9A + 9B = 9(B-A) (mod 101). For this to be 0 (mod 101), A must equal B. Since A cannot be 0, there are 9 choices (1111, 2222, ..., 9999).
A number N, when split into blocks of 3 digits from right to left and summed, gives a multiple of 37. Which must N always be divisible by?
Hint/Explanation: Adding 3-digit blocks tests for divisibility by 999. Since 999 = 27 x 37, a block sum divisible by 37 guarantees the original number is divisible by 37.
Find the remainder when 10¹²- 1 is divided by 1001.
Hint/Explanation: Write 10^12 in terms of 10^3: 10^3 = -1 (mod 1001), so 10^12 = (10^3)^4 = (-1)^4 = 1 (mod 1001). So, 1-1 = 0.
Let X = 555...55 (60 digits) and Y = 999...99 (60 digits). Which of the following primes does not divide X x Y?
Hint/Explanation: 60 is a multiple of 6, so a 60-digit repunit is divisible by 1001 = 7 x 11 x 13. Prime 17 requires a 16-digit-period repunit rule (10^16 = 1 mod 17) and does not divide a 60-digit repunit
Let A be a 50-digit repunit (111...11). Find the remainder when C = A x (10⁵⁰ +1) + 10¹⁰⁰ is divided by 101.
Hint/Explanation: A x (10^50+1) forms a 100-digit rep-unit, which has 50 blocks of 11 and cancels to 0 (mod 101). We are left with 10^100 (mod 101) = (10^2)^50 = (-1)^50 = 1.