Find the smallest 8-digit number which gives 15 as a remainder when divided by 38, 22 and 16. A. 10002004 b. 10002015 c. 10001919 d. 10000015
Answers
Answer:
10001919
Step-by-step explanation:
Let us first find out the LCM of 38,22 and 16.
⇒ LCM(22,38,16)
⇒ 2 × 11 × 19 × 8
= 3344
Hence, every number which is divisible by 3344 is automatically divisible by 38, 22 and 16.
Now we divide 10000000 which is the least number of 8 digits to find out if it is divisible by 3344
We need 10001904 to make is completely divisible by 3344.
Hence, since 15 is the remainder which is left in every case, hence, we can say that 10001904 + 15
Ie 10001919 is the least number which when divided by 38, 22 and 16 leaves a remainder of 15 in every case.
Answer:
The smallest 8-digit number which gives 15 as a remainder when divided by 38, 22 and 16 is 10000015
Step-by-step explanation:
From the above question,
To locate the smallest 8-digit variety that offers a the rest of 15 when divided by using 38, 22, and 16, we can observe these steps:
- Step 1: Find the least frequent more than one (LCM) of the three divisors.
The LCM of 38, 22, and sixteen is 836.
- Step 2: Add the the rest of 15 to the LCM determined in step 1.
836 + 15 = 851.
Therefore,
The smallest 8-digit range that offers a the rest of 15 when divided by means of 38, 22, and sixteen is 10,000,015 (i.e., the 8-digit wide variety that begins with 1 and has 7 zeros after it), seeing that it is the smallest variety that is increased than or equal to 851 and leaves a the rest of 15 when divided by means of 38, 22, and 16.
The smallest 8-digit number which gives 15 as a remainder when divided by 38, 22 and 16 is 10000015
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