Question
Download Solution PDFLet p be the greatest number of 4-digits which when divided by 21, 22, 24 and 28, the remainder in each case is 8. What is the sum of the digits of p ?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
p is the greatest 4-digit number.
When p is divided by 21, 22, 24, and 28, the remainder in each case is 8.
Calculation:
Prime factorization of 21 = 3 × 7
Prime factorization of 22 = 2 × 11
Prime factorization of 24 = 23 × 3
Prime factorization of 28 = 22 × 7
LCM(21, 22, 24, 28) = 23 × 3 × 7 × 11
= 8 × 3 × 7 × 11 = 1848
Now, we need to find the greatest 4-digit number which is of the form 1848 k + 8.
The greatest 4-digit number is 9999.
We need to find the largest 'k' such that 1848k + 8 ≤ 9999
⇒ 1848k ≤ 9999 - 8
⇒ 1848k ≤ 9991
⇒ k ≤ 9991 / 1848
⇒ k ≤ 5.406...
Since k must be an integer, the greatest possible value for k is 5.
Now, substitute k = 5 into the expression for p:
p = 1848 × 5 + 8
⇒ p = 9240 + 8
⇒ p = 9248
Sum of digits of p = 9 + 2 + 4 + 8 = 23
∴ The correct answer is option 4.
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