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Find the least number which when divided by 2, 5, 9 and 12 leaves a remainder 3 but leaves no remainder when same number is divided by 11.
- a)281
- b)357
- c)360
- d)363
- e)224
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Find the least number which when divided by 2, 5, 9 and 12 leaves a re...
LCM(2, 5, 9, 12) = 180
So the number will be somewhat = 180x + 3
This number leaves no remainder when divided by 11, so x = 2 to get (180x+3) fully divided by 11
So number = 180*2 + 3
So the number will be somewhat = 180x + 3
This number leaves no remainder when divided by 11, so x = 2 to get (180x+3) fully divided by 11
So number = 180*2 + 3
Most Upvoted Answer
Find the least number which when divided by 2, 5, 9 and 12 leaves a re...
(USE THIS METHOD ONLY IF ANY ONE OF THE OPTIONS REMAINDERS AS 0 and don't do this method if all the options got the remainder as 0).
So the question was by dividing 11 there will be no remainder. I divided the numbers in the option with 11 and I got remainders as
a) remainder is 6
b)remainder is 5
c)remainder is 8
d) remainder is 0
e) remainder is different
so the option d leaves a remainder is 0 the answer is 363.
I solved this problem by using this method.
So the question was by dividing 11 there will be no remainder. I divided the numbers in the option with 11 and I got remainders as
a) remainder is 6
b)remainder is 5
c)remainder is 8
d) remainder is 0
e) remainder is different
so the option d leaves a remainder is 0 the answer is 363.
I solved this problem by using this method.
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Community Answer
Find the least number which when divided by 2, 5, 9 and 12 leaves a re...
To find the least number that satisfies the given conditions, we need to find the least common multiple (LCM) of the divisors 2, 5, 9, and 12.
Finding the LCM:
Step 1: Prime factorize each number:
2 = 2^1
5 = 5^1
9 = 3^2
12 = 2^2 * 3^1
Step 2: Take the highest power of each prime factor:
2^2 * 3^2 * 5^1 = 4 * 9 * 5 = 180
So, the LCM of 2, 5, 9, and 12 is 180.
Now, we know that when this number is divided by 2, 5, 9, and 12, it leaves a remainder of 3. But we need to find a number that also leaves no remainder when divided by 11.
Finding a number that leaves no remainder when divided by 11:
Step 1: Subtract the remainder (3) from the LCM (180):
180 - 3 = 177
Step 2: Check if the result is divisible by 11:
177 ÷ 11 = 16 with a remainder of 1
Since the result is not divisible by 11, we need to find the next multiple of 180 that is divisible by 11.
Step 3: Add the LCM (180) to the result (177) until we find a number divisible by 11:
177 + 180 = 357
Step 4: Check if the new result is divisible by 11:
357 ÷ 11 = 32 with a remainder of 5
Since 357 is not divisible by 11, we need to find the next multiple of 180 that is divisible by 11.
Step 5: Add the LCM (180) to the result (357) until we find a number divisible by 11:
357 + 180 = 537
Step 6: Check if the new result is divisible by 11:
537 ÷ 11 = 48 with a remainder of 9
Again, 537 is not divisible by 11, so we repeat the process.
Step 7: Add the LCM (180) to the result (537) until we find a number divisible by 11:
537 + 180 = 717
Step 8: Check if the new result is divisible by 11:
717 ÷ 11 = 65 with a remainder of 2
Finally, we have found a number (717) that satisfies all the given conditions. However, we need to find the least number that satisfies the conditions.
Step 9: Continue adding the LCM (180) until we find the least number divisible by 11:
717 + 180 = 897
Step 10: Check if the new result is divisible by 11:
897 ÷ 11 = 81 with a remainder of 6
Continue the process:
897 + 180 = 1077 (not divisible by 11)
1077 + 180 = 1257 (not divisible by 11)
1257 + 180 = 1437 (not divisible by 11)
1437 + 180 = 1617 (not divisible by 11)
1617 + 180 = 1797
Finding the LCM:
Step 1: Prime factorize each number:
2 = 2^1
5 = 5^1
9 = 3^2
12 = 2^2 * 3^1
Step 2: Take the highest power of each prime factor:
2^2 * 3^2 * 5^1 = 4 * 9 * 5 = 180
So, the LCM of 2, 5, 9, and 12 is 180.
Now, we know that when this number is divided by 2, 5, 9, and 12, it leaves a remainder of 3. But we need to find a number that also leaves no remainder when divided by 11.
Finding a number that leaves no remainder when divided by 11:
Step 1: Subtract the remainder (3) from the LCM (180):
180 - 3 = 177
Step 2: Check if the result is divisible by 11:
177 ÷ 11 = 16 with a remainder of 1
Since the result is not divisible by 11, we need to find the next multiple of 180 that is divisible by 11.
Step 3: Add the LCM (180) to the result (177) until we find a number divisible by 11:
177 + 180 = 357
Step 4: Check if the new result is divisible by 11:
357 ÷ 11 = 32 with a remainder of 5
Since 357 is not divisible by 11, we need to find the next multiple of 180 that is divisible by 11.
Step 5: Add the LCM (180) to the result (357) until we find a number divisible by 11:
357 + 180 = 537
Step 6: Check if the new result is divisible by 11:
537 ÷ 11 = 48 with a remainder of 9
Again, 537 is not divisible by 11, so we repeat the process.
Step 7: Add the LCM (180) to the result (537) until we find a number divisible by 11:
537 + 180 = 717
Step 8: Check if the new result is divisible by 11:
717 ÷ 11 = 65 with a remainder of 2
Finally, we have found a number (717) that satisfies all the given conditions. However, we need to find the least number that satisfies the conditions.
Step 9: Continue adding the LCM (180) until we find the least number divisible by 11:
717 + 180 = 897
Step 10: Check if the new result is divisible by 11:
897 ÷ 11 = 81 with a remainder of 6
Continue the process:
897 + 180 = 1077 (not divisible by 11)
1077 + 180 = 1257 (not divisible by 11)
1257 + 180 = 1437 (not divisible by 11)
1437 + 180 = 1617 (not divisible by 11)
1617 + 180 = 1797
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Find the least number which when divided by 2, 5, 9 and 12 leaves a remainder 3 but leaves no remainder when same number is divided by 11.a)281b)357c)360d)363e)224Correct answer is option 'D'. Can you explain this answer?
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