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The largest possible number with which when 38, 66
and 80 are divided, leaves the remainder same is
a)14
b)7
c)28
d)None of these
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The largest possible number with which when 38, 66... more and 80 are ...
Let the common remainder be 10
From the given numbers we now subtract the common remainder
i.e 38 − 10 = 28,66 −10 = 56, 80 −10 = 70
Now we will find the HCF of 28, 56, 70
28 = 2 x 7 x 2
56 = 2 x 2 x 2 x 7
70 = 2 x 5 x 7
So HCF of 28, 56 and 70 = 2 x 7
So the required number is 14
Most Upvoted Answer
The largest possible number with which when 38, 66... more and 80 are ...
Given: Numbers 38, 66, ..., 80
To find: The largest possible number with which when all these numbers are divided, leaves the same remainder.
Solution:
Step 1: Find the difference between consecutive numbers.
66 - 38 = 28
...
80 - 66 = 14
We see that the difference between consecutive numbers is not the same. Therefore, we cannot directly find the required number by taking the LCM of the given numbers.
Step 2: Let us assume that the required number is x.
Then, we can write:
38 ≡ r (mod x)
66 ≡ r (mod x)
...
80 ≡ r (mod x)
Here, 'r' is the common remainder.
Step 3: Subtracting the first two congruences, we get:
28 ≡ 0 (mod x)
Similarly, we can subtract the other congruences to get:
14 ≡ 0 (mod x)
...
80 - 66 ≡ 0 (mod x)
Step 4: Now, we need to find the largest possible number 'x' which divides all these differences (28, 14, ..., 14).
For this, we need to find the GCD of these differences.
GCD(28, 14, 14, 14) = 14
Therefore, the largest possible number 'x' is 14.
Step 5: Finally, we need to check if 14 satisfies the original congruences:
38 ≡ 4 (mod 14)
66 ≡ 12 (mod 14)
...
80 ≡ 8 (mod 14)
We see that 14 satisfies all the congruences. Therefore, the largest possible number with which all the given numbers can be divided, leaving the same remainder, is 14.
Hence, the correct answer is option (a) 14.
To find: The largest possible number with which when all these numbers are divided, leaves the same remainder.
Solution:
Step 1: Find the difference between consecutive numbers.
66 - 38 = 28
...
80 - 66 = 14
We see that the difference between consecutive numbers is not the same. Therefore, we cannot directly find the required number by taking the LCM of the given numbers.
Step 2: Let us assume that the required number is x.
Then, we can write:
38 ≡ r (mod x)
66 ≡ r (mod x)
...
80 ≡ r (mod x)
Here, 'r' is the common remainder.
Step 3: Subtracting the first two congruences, we get:
28 ≡ 0 (mod x)
Similarly, we can subtract the other congruences to get:
14 ≡ 0 (mod x)
...
80 - 66 ≡ 0 (mod x)
Step 4: Now, we need to find the largest possible number 'x' which divides all these differences (28, 14, ..., 14).
For this, we need to find the GCD of these differences.
GCD(28, 14, 14, 14) = 14
Therefore, the largest possible number 'x' is 14.
Step 5: Finally, we need to check if 14 satisfies the original congruences:
38 ≡ 4 (mod 14)
66 ≡ 12 (mod 14)
...
80 ≡ 8 (mod 14)
We see that 14 satisfies all the congruences. Therefore, the largest possible number with which all the given numbers can be divided, leaving the same remainder, is 14.
Hence, the correct answer is option (a) 14.
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Community Answer
The largest possible number with which when 38, 66... more and 80 are ...
Assume that ‘d’ is the largest divisor , which divides 38 , 66 and 80 leaving remainder ‘ r' in each case. given no. = d*q + r , where 0 < r="" />< d="" dividend="divisor" *="" quotient+="" remainder="" 38-r="d" "="" *q1="66-r" ="d"="" *="" q2="" 80-r="d*" q3="" therefore.="" (38-r),(66-r)="&" (80-r)="" are="" exactly="" divisible="" by="" d="" as="" remainder="0." as="" we="" know="" that="" if="" d="" divides="" a="" &="" b.then="" d="" divides="" (a-="" b)="" also.="" so,="" d="" divides="" (66-r)="" -="" (38-r)="d" divides="" 28="" ………..(1)="" similarly="" d="" divides="" (80-="" r)="" -="" (38-r)="d" divides="" 42="" ……….(2)="" from="" (1)="" &="" (2)="" d="" is="" divides="" both="" 28,42="" 28="2*7*2" 42="2*7*3" so,="" hcf="" is="" 14="" ans:="" 14="" d="" dividend="divisor" *="" quotient="" +="" remainder="" 38="d" *="" q1="" +="" r="" 66="d" *="" q2="" +="" r="" 80="d" *="" q3="" +="" r="" or="" 38-r="d" *q1="" 66="" -r="d" *="" q2="" 80="" -r="d*" q3="" therefore.="" (38-r),="" (66-r)="" &="" (80-r)="" are="" exactly="" divisible="" by="" d="" as="" remainder="0." as="" we="" know="" that="" ,="" if="" d="" divides="" a="" &="" b.="" then="" d="" divides="" (a-="" b)="" also.="" so,="" d="" divides="" (66-r)="" -="" (38-r)="d" divides="" 28="" ………..(1)="" similarly="" d="" divides="" (80-="" r)="" -="" (38-r)="d" divides=" 42=" "="" ……….(2)="" from="" (1)="" &="" (2)="" d="" is="" divides="" both="" 28,42="" 28="2*7*2" 42="2*7*3" so,="" hcf="" is="" 14="" ans:="" d="" dividend="divisor" *="" quotient+="" remainder="" 38-r="d" "="" *q1="66-r" ="d"="" *="" q2="" 80-r="d*" q3="" therefore.="" (38-r),(66-r)="&" (80-r)="" are="" exactly="" divisible="" by="" d="" as="" remainder="0." as="" we="" know="" that="" if="" d="" divides="" a="" &="" b.then="" d="" divides="" (a-="" b)="" also.="" so,="" d="" divides="" (66-r)="" -="" (38-r)="d" divides="" 28="" ………..(1)="" similarly="" d="" divides="" (80-="" r)="" -="" (38-r)="d" divides="" 42="" ……….(2)="" from="" (1)="" &="" (2)="" d="" is="" divides="" both="" 28,42="" 28="2*7*2" 42="2*7*3" so,="" hcf="" is="" 14="" ans:="" 14="" d="" dividend="divisor" *="" quotient="" +="" remainder="" 38="d" *="" q1="" +="" r="" 66="d" *="" q2="" +="" r="" 80="d" *="" q3="" +="" r="" or="" 38-r="d" *q1="" 66="" -r="d" *="" q2="" 80="" -r="d*" q3="" therefore.="" (38-r),="" (66-r)="" &="" (80-r)="" are="" exactly="" divisible="" by="" d="" as="" remainder="0." as="" we="" know="" that="" ,="" if="" d="" divides="" a="" &="" b.="" then="" d="" divides="" (a-="" b)="" also.="" so,="" d="" divides="" (66-r)="" -="" (38-r)="d" divides="" 28="" ………..(1)="" similarly="" d="" divides="" (80-="" r)="" -="" (38-r)="d" divides=" 42=" "="" ……….(2)="" from="" (1)="" &="" (2)="" d="" is="" divides="" both="" 28,42="" 28="2*7*2" 42="2*7*3" so,="" hcf="" is="" 14="" ans:="" />
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The largest possible number with which when 38, 66... more and 80 are divided, leaves the remainder same is a)14b)7c)28d)None of theseCorrect answer is option 'A'. Can you explain this answer?
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