LCM & HCF - 2 | Algebra - Number Theory
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What is the least number that must be subtracted from 1850, so that the remainder when divided by 7, 12, and 16 is 4?
- (a)
131
- (b)
1355
- (c)
134
- (d)
166
Answer: Option D
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Explanation :
When a number N is divided by 7, 12 and 16, the remainder is 4.
⇒ N = k × LCM (7, 12, 16) + 4, where k = 1, 2, 3… and so on.
⇒ N = 336k + 4
When k = 5, N = 1684.
So, the number to be subtracted from 1850 is = 1850 – 1684 = 166
Hence, option (d).
Workspace:
What is the smallest number which when divided by 7, 9, and 11 leaves remainders 5, 7, and 9 respectively?
- (a)
1265
- (b)
691
- (c)
541
- (d)
1013
- (e)
987
Answer: Option B
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Explanation :
Let the required number be x.
Here divisors are 7, 9, 11 and remainders are 5, 7, 9.
The difference between respective divisors and remainders is 2.
∴ x = k × LCM (7, 9, 11) – 2 = 693k – 2.
x will be smallest when k = 1.
⇒ smallest value of x = 693 × 1 – 2 = 691.
Hence, option (b).
Workspace:
A number leaves a remainder of 2 and 4 respectively when divided by 4 and 5. Find the least three digit such number.
Answer: 114
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Explanation :
Let the number be N.
∴ N = 4a + 2 i.e., N can be 2, 6, 10, 14, 18, ….
Also, N = 5b + 4 i.e., N can be 4, 9, 14, 19, …
The least possible value of N = 14.
∴ N = k × LCM(4, 5) + 14 = 20k + 14.
Now we have to find the least three-digit such number.
For k = 5, N = 114 (least three-digit such number).
Hence, 114.
Workspace:
Find the greatest number which when divides 252, 1848, and 462 leaves the same remainder.
- (a)
64
- (b)
42
- (c)
36
- (d)
21
- (e)
84
Answer: Option B
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Explanation :
The greatest number which when divides a, b, and c leaves the same remainder = HCF [|a – b|, |b – c|].
The greatest number which when divides 252, 1848, and 462 leaves the same remainder
= HCF ((1848 – 252), (1848 – 462)) = 42.
Hence, option (b).
Workspace:
Find the least number which when divided by 6, 15, 17 leaves a remainder 1, but when divided by 7 leaves no remainder.
Answer: 511
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Explanation :
Any number, which when divided by 6, 15 and 17 leaves a remainder of 1, is of the form {k × LCM(6, 15, 17) + 1}, where = 1, 2, 3… and so on.
k × LCM(6, 15, 17) + 1 = 510k + 1.
It is given that (510k + 1) is a multiple of 7.
When k = 1, we get 511, which is a multiple of 7.
Hence, 511.
Workspace:
Find the HCF and LCM of the polynomials (y2 - 5y + 6) and (y2 - y - 2).
- (a)
(y - 2), (y - 2) (y - 3) (y + 1)
- (b)
(y - 2), (y - 2) (y - 3)
- (c)
(y - 3), (y - 2) (y - 3) (y - 1)
- (d)
(y – 3), (y + 2) (y - 3) (y - 1)
- (e)
(y - 2), (y - 2)
Answer: Option A
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Explanation :
(y2 - 5y + 6) and (y2 - 7y + 10)
y2 - 5y + 6 = (y2 - 3y - 2y + 6) = (y2 - 3y) - (2y - 6) = y(y - 3) - 2(y - 3) = (y - 3) (y - 2)
y2 – y - 2 = (y2 - 2y + y - 2) = (y2 - 2y) + (y - 2) = y(y - 2) + (y - 2) = (y + 1)(y - 2)
LCM = (y - 2) (y - 3) (y + 1)
HCF = (y - 2)
Hence, option (a).
Workspace:
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