WebLet 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: A. 4 B. 5 C. 6 D. 8 Asked In CSC … WebApr 7, 2024 · The given numbers are 1305, 4665 and 6905. We have to find the greatest number N that will divide 1305, 4665 and 6905 leaving the same remainder. The greatest …
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WebWhat is the HCF of a 2 b 4 + 2 a 2 b 2 and (ab) 7 - 4a 2 b 2? Options A. ab B. a2 b3 C. a2 b2 D. a3 b2 Show Answer Scratch Pad Discuss 7. Find the largest number which divides 1305, 4665 and 6905 leaving same remainder in each case. Also, find the common Remainder. Options A. 1210, 158 B. 1120, 158 C. 1120, 185 D. 1210, 185 Show Answer … WebN = H.C.F. of (6905 - 4665), (4665 - 1305), and (6905 - 1305) = H.C.F. of 2240, 3360 and 5600 . HCF of 5600 and 3360: 5600 = 3360 × 1 + 2240 3360 = 2240 × 1 + 1120 2240 = …
Webarithmetic aptitude :: hcf and lcm :: general questions. hcf and lcm - General Questions; hcf and lcm - Important Formulas and Facts ... 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 : A) 4. B) 5. C) 6. D) 8. View Answer; Report; Discuss WebA. 4 B. 3 C. 6 D. 5 Solution: If the remainder is same in each case and remainder is not given, HCF of the differences of the numbers is the required greatest number 6905 - 1305 = 5600 6905 - 4665 = 2240 4665 - 1305 = 3360 Hence, the greatest number which divides 1305, 4665 and 6905 and gives the same remainder, N = HCF of 5600, 2240, 3360 ...
WebJun 22, 2024 · Aishwarya06. Since the remainder is same in each case, we will use the following formula-. H.C.F (x, y, z)= H.C.F of (x -y), (y- z), ( z-x) So, let's take out the … WebJun 22, 2024 · So, let's take out the H.C.F of 1305, 4665 and 6905 (leaving the same remainder) = H.C.F of (4665 - 1305), ( 6905- 4665) and ( 6905- 1305) = H.c.f of 3360, 2240 and 5600 which is= 1120. Sum of digits of 1120= (1+1+2+0)= 4. Answer=4. Find Math textbook solutions? Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 …
WebStep 1: Find the prime factors for 40 and 60, The prime factorization of 40 is 2 x 5. The prime factorization of 60 is 2 x 3 x 5. Step 2: List out the highest number of common prime factors of 40 and 60 ie., 2 x 2 x 5. Step 3: Now, on multiplying the common prime factors we will get the HCF of two numbers. 2 * 2 * 5 = 40.
Web135 = 3 × 3 × 3 × 5. Find the prime factorization of 405. 405 = 3 × 3 × 3 × 3 × 5. To find the GCF, multiply all the prime factors common to both numbers: Therefore, GCF = 3 × 3 × 3 … data:image/svg+xml charset us-asciiWeb1) 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 : A) 4 B) 5 C) 6 D) 8 Answer: Option A Explanation: N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305) = H.C.F. of 3360, 2240 and 5600 = 1120. Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4 Note: martinelli giovanni ematologoWebThe 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, … data image processingWebMar 15, 2024 · The HCF and LCM of the two numbers are 13 and 455 respectively. If one of the numbers lies between 75 and 125, then, that numbers are: (a) 78 (b) 91 (c) 104 (d) 117. ... Let N be the greatest number that will divide 1305, 4665, and 6905 leaving the same remainder in each case. Then, the sum of the digits in N is: (a) 4 (b) 5 (c) 6 data image touch encoderWeb(4665 - 1305)=3360, (6905 - 4665)=2240 and (6905 - 1305)=5600 3360:2240:5600 1120:1120:1120 Let the numbers be represented as . 1305 = Na + d -----(1) where q is the remainder ... Now we need to find HCF of (3360 , 5600 , 2240) , that is 80 which is equal to value of N. Thus N is 1120. Suggest Corrections. 0. Similar questions. data:image/svg+xml utf8 svghttp://placementstudy.com/arithmetic-aptitude/3/hcf-and-lcm/43 data imaging solutionsWebHCF & LCM. 211 . If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to: ... 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: A. 4. B. 5. C. 6. D. 8 ... martinelli giovanna parma