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ab and cd are 2 two-digit natural numbers. Given that 4b + a = 13k1 and 5d - c = 17k2, where k1 and k2 are natural numbers. Find the largest number that will always divide the product of ab and cd.
- a)144
- b)169
- c)221
- d)256
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
ab and cd are 2 two-digit natural numbers. Given that 4b + a = 13k1and...
To find the largest number that will always divide the product of ab and cd, we need to find the common factors of ab and cd.
Given that 4b - a = 13k1 and 5d - c = 17k2, where k1 and k2 are natural numbers, let's solve these equations to get some insights.
Solving the first equation, we get:
4b - a = 13k1
=> a = 4b - 13k1
Solving the second equation, we get:
5d - c = 17k2
=> c = 5d - 17k2
Now, let's substitute the values of a and c in the product ab * cd:
(ab) * (cd) = (4b - 13k1) * (5d - 17k2)
Expanding the above expression, we get:
(ab) * (cd) = 20bd - 68b * k2 - 65d * k1 + 221k1 * k2
From the above expression, we can observe that the product ab * cd is divisible by 221, as 221 appears as a common factor.
Therefore, the largest number that will always divide the product of ab and cd is 221.
Hence, the correct answer is option C) 221.
Given that 4b - a = 13k1 and 5d - c = 17k2, where k1 and k2 are natural numbers, let's solve these equations to get some insights.
Solving the first equation, we get:
4b - a = 13k1
=> a = 4b - 13k1
Solving the second equation, we get:
5d - c = 17k2
=> c = 5d - 17k2
Now, let's substitute the values of a and c in the product ab * cd:
(ab) * (cd) = (4b - 13k1) * (5d - 17k2)
Expanding the above expression, we get:
(ab) * (cd) = 20bd - 68b * k2 - 65d * k1 + 221k1 * k2
From the above expression, we can observe that the product ab * cd is divisible by 221, as 221 appears as a common factor.
Therefore, the largest number that will always divide the product of ab and cd is 221.
Hence, the correct answer is option C) 221.
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Community Answer
ab and cd are 2 two-digit natural numbers. Given that 4b + a = 13k1and...
ab × cd = (10a + b)(10c + d)
= (40b + 10a - 39b)(51d - (50d - 10c))
= (10(4b + a) - 39b)(51d - 10(5d - c))
= (10 × 13k1 - 39b)(51d - 10 × 17k2)
= 13 × 17(10k1 - 3b)(3d - 10k2)
= 221(10k1 - 3b)(3d - 10k2)
Hence, the largest number that will divide the product ab and cd is 221.
= (40b + 10a - 39b)(51d - (50d - 10c))
= (10(4b + a) - 39b)(51d - 10(5d - c))
= (10 × 13k1 - 39b)(51d - 10 × 17k2)
= 13 × 17(10k1 - 3b)(3d - 10k2)
= 221(10k1 - 3b)(3d - 10k2)
Hence, the largest number that will divide the product ab and cd is 221.
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ab and cd are 2 two-digit natural numbers. Given that 4b + a = 13k1and 5d - c = 17k2, where k1and k2are natural numbers. Find the largest number that will always divide the product of ab and cd.a)144b)169c)221d)256Correct answer is option 'C'. Can you explain this answer?
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