DSSSB TGT CS 2023 Shift 2 – Question 042

Question:

Let \(x\) be the least number which when divided by 11, 21, 24 and 28, leaves a remainder of 10 in each case. If \(x\) is divisible by 37, what will be the sum of the digits of \(x\)?

मान लीजिए \(x\) वह सबसे छोटी संख्या है जिसे 11, 21, 24 और 28 से भाग देने पर प्रत्येक स्थिति में शेषफल 10 बचता है। यदि \(x\), 37 से विभाज्य है, तो \(x\) के अंकों का योग क्या होगा?

Options:

  • 13
  • 15
  • 16
  • 17
  • 13
  • 15
  • 16
  • 17

Correct Answer: c)

Explanation:

Step 1: Find the LCM of the divisors (11, 21, 24, 28):
Prime factorizations:
\[ 11 = 11^1 \]
\[ 21 = 3 \times 7 \]
\[ 24 = 2^3 \times 3 \]
\[ 28 = 2^2 \times 7 \]
\[ \text{LCM}(11, 21, 24, 28) = 2^3 \times 3^1 \times 7^1 \times 11^1 = 8 \times 21 \times 11 = 168 \times 11 = 1848 \]

Step 2: General form of the number:
Any number leaving a remainder of 10 when divided by 11, 21, 24, and 28 can be written as:
\[ x = 1848k + 10 \quad (\text{where } k \ge 1 \text{ is an integer}) \]

Step 3: Apply the condition of divisibility by 37:
Divide 1848 by 37:
\[ 1848 = 37 \times 49 + 35 \implies 1848 \equiv 35 \equiv -2 \pmod{37} \]
Therefore:
\[ x \equiv (-2k + 10) \equiv 0 \pmod{37} \]
\[ 2k \equiv 10 \pmod{37} \implies k \equiv 5 \pmod{37} \]
For the least number \(x\), choose \(k = 5\).

Step 4: Calculate \(x\) and the sum of its digits:
\[ x = 1848 \times 5 + 10 = 9240 + 10 = 9250 \]
Check divisibility: \(9250 \div 37 = 250\) (completely divisible).
Sum of digits of \(x\):
\[ 9 + 2 + 5 + 0 = 16 \]
Hence, the correct answer is option c).

चरण 1: भाजकों (11, 21, 24, 28) का LCM ज्ञात करें:
अभाज्य गुणनखंडन:
\[ 11 = 11 \]
\[ 21 = 3 \times 7 \]
\[ 24 = 2^3 \times 3 \]
\[ 28 = 2^2 \times 7 \]
\[ \text{LCM}(11, 21, 24, 28) = 8 \times 3 \times 7 \times 11 = 1848 \]

चरण 2: संख्या का सामान्य रूप:
प्रत्येक स्थिति में 10 शेषफल देने वाली संख्या:
\[ x = 1848k + 10 \quad (\text{जहाँ } k \text{ एक धनात्मक पूर्णांक है}) \]

चरण 3: 37 से विभाज्यता की शर्त लागू करें:
1848 को 37 से भाग देने पर:
\[ 1848 = 37 \times 49 + 35 \equiv -2 \pmod{37} \]
अतः:
\[ -2k + 10 \equiv 0 \pmod{37} \implies 2k \equiv 10 \pmod{37} \implies k \equiv 5 \pmod{37} \]
न्यूनतम मान के लिए \(k = 5\) लेते हैं।

चरण 4: \(x\) और उसके अंकों का योग ज्ञात करें:
\[ x = 1848 \times 5 + 10 = 9240 + 10 = 9250 \]
जाँच: \(9250 \div 37 = 250\) (पूर्णतः विभाज्य)।
अंकों का योग:
\[ 9 + 2 + 5 + 0 = 16 \]
अतः सही उत्तर विकल्प c) है।

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top