CUET UG 2025 — General Test Quantitative Reasoning

Take the number 2345

Sum of digits:

$$2 + 3 + 4 + 5 = 14$$2+3+4+5=14

Subtract from the original number:

$$2345 - 14 = 2331$$2345−14=2331

$2331 \div 9 = 259$2331÷9=259 with no remainder

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Take the number 1678

Sum of digits:

$$1 + 6 + 7 + 8 = 22$$1+6+7+8=22

Subtract from the original number:

$$1678 - 22 = 1656$$1678−22=1656

$1656 \div 9 = 184$1656÷9=184 with no remainder

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Let the four digits be $a, b, c, d$a,b,c,d where $a$a is the thousands digit, $b$b is the hundreds digit, $c$c is the tens digit, and $d$d is the ones digit.

The 4-digit number can be written as:

$$1000a + 100b + 10c + d$$1000a+100b+10c+d

Sum of digits:

$$a + b + c + d$$a+b+c+d

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Number minus sum of digits:

$$(1000a + 100b + 10c + d) - (a + b + c + d)$$(1000a+100b+10c+d)−(a+b+c+d)

$$= 1000a - a + 100b - b + 10c - c + d - d$$=1000a−a+100b−b+10c−c+d−d

$$= 999a + 99b + 9c$$=999a+99b+9c

$$= 9(111a + 11b + c)$$=9(111a+11b+c)

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The result is $9(111a + 11b + c)$9(111a+11b+c), which is always a multiple of 9.

Therefore, the resulting number is always divisible by 9.