LCM and HCF
- Which greatest number will divide 3026 and 5053 leaving remainders 11 and 13 respectively?
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We know that the largest number which when divide the numbers a, b and c give remainders as p, q, r respectively is given by H.C.F. of (a – p), (b – q) and (c – r).
3026 –11 = 3015 and 5053 –13 = 5040
Required number = HCF of 3015 and 5040
Correct Option: C
We know that the largest number which when divide the numbers a, b and c give remainders as p, q, r respectively is given by H.C.F. of (a – p), (b – q) and (c – r).
3026 –11 = 3015 and 5053 –13 = 5040
Required number = HCF of 3015 and 5040
∴ Required number = HCF of 3015 and 5040 = 45
- What is the greatest number that will divide 307 and 330 leaving remainders 3 and 7 respectively ?
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As we know that the largest number which when divide the numbers a, b and c give remainders as p, q, r respectively is given by H.C.F. of (a – p), (b – q) and (c – r).
The number will be HCF of 307 – 3 = 304 and 330 – 7 = 323.
Correct Option: A
As we know that the largest number which when divide the numbers a, b and c give remainders as p, q, r respectively is given by H.C.F. of (a – p), (b – q) and (c – r).
The number will be HCF of 307 – 3 = 304 and 330 – 7 = 323.
∴ Required number = 19
- Let N be the greatest number that will divide 1305, 4665 and 6905 leaving the same remainder in each case. Then, sum of the digits in N is :
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We can say that the largest number which when divide the numbers a, b and c give remainders as p, q, r respectively is given by H.C.F. of (a – p), (b – q) and (c – r).
The greatest number N = HCF of (1305 – t ), (4665 – t ) and (6905 – t), where t is the remainder
= HCF of (4665 – 1305), (6905– 4665) and (6905 – 1305)
= HCF of 3360, 2240 and 5600
Correct Option: A
We can say that the largest number which when divide the numbers a, b and c give remainders as p, q, r respectively is given by H.C.F. of (a – p), (b – q) and (c – r).
The greatest number N = HCF of (1305 – t ), (4665 – t ) and (6905 – t), where t is the remainder
= HCF of (4665 – 1305), (6905– 4665) and (6905 – 1305)
= HCF of 3360, 2240 and 5600
∴ N = 1120
Sum of digits = 1 + 1 + 2 + 0 = 4
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H.C.F of 2 , 4 and 6 is 3 5 7
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H.C.F of 2 , 4 and 6 3 5 7
As we know that ,H.C.F. of fractions = HCF of numerators LCM of denominators
Correct Option: B
H.C.F of 2 , 4 and 6 3 5 7
As we know that ,H.C.F. of fractions = HCF of numerators LCM of denominators H.C.F. = HCF of 2, 4 and 6 LCM of 3, 5 and 7 H.C.F. = 2 105
- The greatest number, which when divide 989 and 1327 leave remainders 5 and 7 respectively, is :
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We know that The largest number which when divide the numbers a, b and c give remainders as p, q, r respectively is given by H.C.F. of (a – p), (b – q) and (c – r).
Required number = HCF of (989 – 5) and (1327 – 7)
Required number = HCF of 984 and 1320Correct Option: C
We know that The largest number which when divide the numbers a, b and c give remainders as p, q, r respectively is given by H.C.F. of (a – p), (b – q) and (c – r).
Required number = HCF of (989 – 5) and (1327 – 7)
Required number = HCF of 984 and 1320 = 24
∴ HCF = 24
