Answer
Hint: When the required number divides the given numbers (285 and 1249), it leaves certain remainders, thus the difference between the two will provide us those numbers that are completely divisible by the required number.
Complete step-by-step answer:
As we have to find the greatest number that divides the two, we need to highest common factor (H.C.F) of the two numbers obtained after subtraction.The numbers are given by the difference exactly divisible by the required number are given as:285 – 9 = 2761249 – 7 = 1242To find the greatest number that exactly divides these are given by their H.C.F:276 = 2 X 2 X 3 X 231242 = 2 X 3 X 3 X 3 X 23Common numbers = 2 X 3 X 23Highest Common Factor (H.C.F) = 6 X 23Highest Common Factor (H.C.F) = 138Therefore, the greatest number which divides 285 and 1249 leaving the remainder 9 and 7 respectively is 138, option D).So, the correct answer is “Option D”.
Note: A factor is a number that divides another number without leaving any remainder.
H.C.F of two numbers is the product of common prime factors between the two where a prime number is divisible only by 1 and itself.H.C.F of the two numbers cannot be greater than either of them.H.C.F can also be called G.C.F which stands for greatest common factor.
TCS Programming and Technical Numerical Ability LCM and HCF
- Hcf ( (148-4), (246-6), (623-11))=12
- 6 years agoHelpfull: Yes(9) No(5)
- 148-4=144
246-6=240
623-11=612
Hence these 3 are divided by 12 leaving remainders 4 6 and 11
- 6 years agoHelpfull: Yes(6) No(0)
- hcf of 144, 240,612 is 12. 12 is the answer
- 6 years agoHelpfull: Yes(2) No(1)
- difference of the dividend and remainder gives you the divisible number
so Hcf of ((148-4),(246-6),(623-11)
144=12*12
240=12*20
623=12*51
so the HCF is 12.
- 5 years agoHelpfull: Yes(2) No(0)
- hcf of given numbers is 12
- 6 years agoHelpfull: Yes(0) No(0)
- a*4/100=8 , a=800/4, a=200
b*8/100 , b=400/8, b=50
c=50/200, c=1/4
- 5 years agoHelpfull: Yes(0) No(2)
- 148-4=144
246-6=240
623-11=612
Hence these 3 are divided by 12 leaving remainders 4 6 and 11
- 3 years agoHelpfull: Yes(0) No(0)
- Ans2 148-4=144
246-6=240
623-11=612 144-= 2*2*18
240= 2*2*3*20
612=2*2*
2 great no
- 3 years agoHelpfull: Yes(0) No(0)
- We know that When a greatest number that divide x,y,z and leaves remainder a,b,c respectively = HCF of (x - a),(y - b),(z - c). Now, HCF of (148 - 4),(246 - 6),(623 - 11) = > HCF of (144, 240,612) Prime factorization of 144 = 2 * 2 * 2 * 2 * 3 * 3 Prime factorization of 240 = 2 * 2 * 2 * 2 * 3 * 5 Prime factorization of 612 = 2 * 2 * 3 * 3 * 17 HCF(144,240,612) = 2 * 3 * 3 = 12. Hence, the greatest number is n = 12. Now, We have to find the remainder when n is divided by 7. = > 7)12(1 7 --- 5.
Therefore, the remainder is 5.
- 3 years agoHelpfull: Yes(0) No(0)
- 148-4=144,246-6=240,623-11=612
HCF of 144,240,612 is 12. - 2 years agoHelpfull: Yes(0) No(0)
Find the greatest number that will divide 148 246 and 623 leaving remainders 4 6 and 11 respectively ?
Remainder in case of 148 = 4 Number = 148 – 4 = 144 Remainder in case of 246 = 6 Number = 246 – 6 = 240 Remainder in case of 623 = 11 Number = 623 – 11 = 612
Now, we have find the H.C.F. of 144, 240 and 612
H.C.F. of 144, 240 and 612 = 2 * 2 * 3 = 12
Therefore, the required number is 12 that will divide 148, 246 and 623 leaving remainders 4, 6 and 11 respectively.
- 24 and 48 leaves remainder 33 21 and 45 respectively,
- 246,
- 246 and 623 leaving remainders 4,
- 6 and 11 respectively,
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