Let x be a least number which when divided by 21,33,35 and 55 leaves in each case a remainder 3, but is exactly divisible by 67. The sum of digits of
Answers
Answer:
x = 4623
Step-by-step explanation:
Let x be a least number which when divided by 21,33,35 and 55 leaves in each case a remainder 3, but is exactly divisible by 67. Find the number
Let say number = x
21a = x -3
33b = x -3
35c = x - 3
55d = x -3
67e = x
21 a = 33b = 35c = 55d
3*7a = 3 * 11b = 5*7c = 5 * 11d = 3*7*11*5*y
a = 5*11y , b = 7*5y , c = 3*11y , d = 3*7y
a = 55y , b = 35y c = 33y d = 21y
21a = 33b = 35c = 55d = 1155y
x - 3 = 1155y
x = 1155y + 3
67e = 1155y + 3
67e = (1139y + 16y) + 3
=> 67e = 67*17y + 16y + 3
16y + 3 should be equal to 67 or multiple of 67
y = 4 will make equation true
=>16*4 + 3 = 67
x = 1155*4 + 3
x = 4623
Verification
4623/21 = 220 * 21 + 3
4623/33 = 140 * 33 + 3
4623/35 = 132 * 35 + 3
4623/55 = 84 * 55 + 3
4623/67 = 69
Answer:
15
Step-by-step explanation:
A/Q
x= 21 k1 +3
x =33 k2 +3
x =35 k3 +3
x =55 k4 +3
x =67 k5
= LCM (211 331 351 55)k + 3
= 1155k + 3
A/Q x = 67 k5
67 k5= 1155 k + 3
at k = 4
Number = 4623 which
satisfies all conditions.
sum of digits = 4 + 6 + 2 + 3 = 15