How many 2-digit positive integers are divisible by 4 or 9?
A. 32 B. 22 C. 30 D. 34 Answer: Option CShow Answer
Solution(By Apex Team)
Number of 2-digit positive integers divisible by 4
The smallest 2-digit positive integer divisible by 4 is 12. The largest 2-digit positive integer divisible by 4 is 96.
All the 2-digit positive integers are terms of an Arithmetic progression with 12 being the first term and 96 being the last term.
The common difference is 4.
The nth term $a_{n}=a_{1}+(n-1) d$, where a1 is the first term, ‘n’ number of terms and ‘d’ the common difference.
So, 96 = 12 + (n – 1) × 4
84 = (n – 1) × 4
Or (n – 1) = 21
Hence, n = 22
i.e., there are 22, 2-digit positive integers that are divisible by 4.
Number of 2-digit positive integers divisible by 9 The smallest 2-digit positive integer divisible by 9 is 18. The largest 2-digit positive integer divisible by 9 is 99. All the 2-digit positive integers are terms of an Arithmetic progression with 18 being the first term and 99 being the last term. The common difference is 9 The nth term $a_{n}=a_{1}+(n-1) d$, where a1 is the first term, ‘n’ number of terms and ‘d’ the common difference. So, 99 = 18 + (n – 1) × 9 Or 81 = (n – 1) × 9 Or (n – 1) = 9 Hence, n = 10 i.e., there are 10 2-digit positive integers that are divisible by 9. Removing double count of numbers divisible by 4 and 9 Numbers such as 36 and 72 are multiples of both 4 and 9 and have therefore been counted in both the groups. There are 2 such numbers. Hence, number of 2-digit positive integers divisible by 4 or 9 = Number of 2-digit positive integers divisible by 4 + Number of 2-digit positive integers divisible by 4 – Number of 2-digit positive integers divisible by 4 and 9 = 22 + 10 – 2 = 30
Number of 2-digit positive integers divisible by 9 The smallest 2-digit positive integer divisible by 9 is 18. The largest 2-digit positive integer divisible by 9 is 99. All the 2-digit positive integers are terms of an Arithmetic progression with 18 being the first term and 99 being the last term. The common difference is 9 The nth term $a_{n}=a_{1}+(n-1) d$, where a1 is the first term, ‘n’ number of terms and ‘d’ the common difference. So, 99 = 18 + (n – 1) × 9 Or 81 = (n – 1) × 9 Or (n – 1) = 9 Hence, n = 10 i.e., there are 10 2-digit positive integers that are divisible by 9. Removing double count of numbers divisible by 4 and 9 Numbers such as 36 and 72 are multiples of both 4 and 9 and have therefore been counted in both the groups. There are 2 such numbers. Hence, number of 2-digit positive integers divisible by 4 or 9 = Number of 2-digit positive integers divisible by 4 + Number of 2-digit positive integers divisible by 4 – Number of 2-digit positive integers divisible by 4 and 9 = 22 + 10 – 2 = 30
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