Shortcut Trick
For a number to be divisible by 2, 5, and 10, its last digit must be 0.
Only the numbers 589270 and 1234560 end in 0.
Now check divisibility by 11 (difference between the sum of alternate digits must be 0 or a multiple of 11):
For 589270: (5 + 9 + 7) − (8 + 2 + 0) = 21 − 10 = 11 (Divisible by 11).
∴ The correct answer is 589270.
Alternate Method
Given:
Numbers: 203467, 830942, 589270, 1234560
Divisors: 2, 5, 10, and 11
Method Used:
Divisibility by 10: The unit digit must be 0.
Divisibility by 11: The difference between the sum of digits at odd positions and even positions must be 0 or a multiple of 11.
Calculations:
⇒ Step 1: Check divisibility by 10.
⇒ 203467 ends in 7 → Not divisible.
⇒ 830942 ends in 2 → Not divisible.
⇒ 589270 and 1234560 end in 0 → Divisible by 2, 5, and 10.
⇒ Step 2: Check divisibility by 11.
⇒ For 589270: Sum of odd place digits = 5 + 9 + 7 = 21.
⇒ Sum of even place digits = 8 + 2 + 0 = 10.
⇒ Difference = 21 − 10 = 11 (which is a multiple of 11).
⇒ For 1234560: Sum of odd place digits = 1 + 3 + 5 + 0 = 9.
⇒ Sum of even place digits = 2 + 4 + 6 = 12.
⇒ Difference = 12 − 9 = 3 (not a multiple of 11).
⇒ Therefore, 589270 is divisible by all the given numbers.
∴ The correct answer is 589270.
Additional Information
Divisibility Rule of 3 and 9
A number is divisible by 3 or 9 if the sum of all its digits is divisible by 3 or 9 respectively.
Divisibility Rule of 4 and 8
A number is divisible by 4 if its last two digits form a number divisible by 4. It is divisible by 8 if its last three digits form a number divisible by 8.
Divisibility Rule of 7 and 13
Group the digits in blocks of 3 from right to left. If the difference of the alternating sum of these blocks is divisible by 7 or 13, the entire number is respectively divisible.
LCM of Divisors
If a number is divisible by a set of co-prime numbers, it must be divisible by their LCM. For example, LCM(2, 5, 10, 11) = 110, so any number divisible by all must be a multiple of 110.