find the smallest number which one divided by 21, 45and 36 will leave a reminder of 7 in each case
Answers
Answer:
9
But if we take as granted that the divisors must be smaller than the divisitor then:
First you find the smallest number that can be divided by 21, 28, 36 and 45.
21: 3×7
28: 2²×7
36: 2²x3²
45: 3²×5
or
21 28 36 45 |2
21 14 18 45 |2
21 7 9 45 |3
7 7 3 15 |3
7 7 1 5 |5
7 7 1 1 |7
1 1 1 1
You conclude on this: 2²×3²×5×7=1260
Then you just add 9 to the number you found: 1269
100% guaranteed placement.
Let X be the smallest number with the desired properties.
Thus X−9 is the smallest integer divisible by 21,28 36,45
Which implies that X−9 is the lcm of 21,28,36 and 45.
LCM have a distributive property
LCM(21,28,36,45)=LCM(LCM(21,28),LCM(36,45))=LCM(84,180)=1260
Thus
X=1269
21=3×728=22×736=22×3245=32×5
L.C.M. (21, 28, 36, 45)=22×32×5×7=1260
1260+9=1269
∴1269 is the smallest number which when divided by 21, 28, 36 and 45
leaves a remainder of 9.
How do you find smallest number which leaves remainder of 9 in each case when divided by 21, 28 , 36 , 45?
A number b is such that when it is divided by 27,30 or 45 , the remainder is 3. What is the smallest possible value of n?
What is the smallest number that when divided by 35, 36, and 91 leaves a remainder of 3?
Find the lowest common multiple of 21,28,36 & 45 and then add 9.
7 / 21 , 28 , 36 , 45
3 / 3 , 4 , 12 , 15
4 / 3 , 1 , 3 , 15
3 / 1 , 1 , 1 , 5
7 x 3 x 4 x 3 x 5 = 1,260
1,260 + 9 = 1,269
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THE PRIME FACTORS OF 21,18,36,45 ARE
21=3*7
28=2*2*7
36=2*2*3*3
45=3*3*5
So LCM( 21,28,36,45)=2*2*3*3*5*7=1260
The remainder in all cases is 9
thus the number=1260+9=1269
21=3×7
28=2×2×7
36= 2×2×3×3
45= 3×3×5
LCM OF (21,28,36,45) = 3780
Now we wants to remainder 9 then we add 9 into the 3780
3780+9= 3789 answer
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3 is the answer sorry that time i would not abe to exolain