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06 Jan 2025, 03:40
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Hi,
I also had similar doubt before. Let me explain how it works.
Series with 1 as remainder after dividing by 7 = 8, 15, 22, 29, 36,.....
Now, if we see among the above series which numbers give 2 as a remainder, we can get the below series.
Dividing the above series by 3 and see if the remainder is 2 = R(8/3)=2, R(15/3)=0, R(22/3) =1, R(29/3) = 2.....
Note in R(8/3)=2, 2 is the remainder when 8 is Divided by 3
So the remainder cycle is = 2, 0, 1, 2, 0, 1.....
Note: The first number of the cycle gives the remainder 2 (it is important, I will explain why)
Since we know there are 71 numbers below 500 that are divisible by 7. 499/7 = 71.
We want a number that will give 1 as the remainder after dividing by 7.
So the first number in our series will be 8, and the last will be 498.
And we already know the remainder cycle is 3. So we need to divide the number 71/3= 23.66
Now, as we know, up to 69th number among 71, there will be 69/3 =23 numbers.
But in the remaining 70th and 71st numbers, the 70th number will also give remainder 2 since our cycle starts with remainder 2 (as I mentioned before)
So the extra number is coming from 70th number from the series.
So total number = 23 +1 = 24
I hope it helps.
I also had similar doubt before. Let me explain how it works.
Series with 1 as remainder after dividing by 7 = 8, 15, 22, 29, 36,.....
Now, if we see among the above series which numbers give 2 as a remainder, we can get the below series.
Dividing the above series by 3 and see if the remainder is 2 = R(8/3)=2, R(15/3)=0, R(22/3) =1, R(29/3) = 2.....
Note in R(8/3)=2, 2 is the remainder when 8 is Divided by 3
So the remainder cycle is = 2, 0, 1, 2, 0, 1.....
Note: The first number of the cycle gives the remainder 2 (it is important, I will explain why)
Since we know there are 71 numbers below 500 that are divisible by 7. 499/7 = 71.
We want a number that will give 1 as the remainder after dividing by 7.
So the first number in our series will be 8, and the last will be 498.
And we already know the remainder cycle is 3. So we need to divide the number 71/3= 23.66
Now, as we know, up to 69th number among 71, there will be 69/3 =23 numbers.
But in the remaining 70th and 71st numbers, the 70th number will also give remainder 2 since our cycle starts with remainder 2 (as I mentioned before)
So the extra number is coming from 70th number from the series.
So total number = 23 +1 = 24
I hope it helps.
valeriugirlea
09 Oct 2025, 02:40
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I used a different approach can some let me know if this is one of the quickest ways to solve-
1) The first minimum common number which will satisfy the condition is 8.
2) After that the second largest will be 29(8 + LCM of 7 and 3).
3) Now according to the condition the integers are less than 500 and the numbers after 29 will be in a A.P, so I used an options and used the A.P formula.
4) So, the option C 24, in the formula will be a21 = 8+23(21), common difference is 21 and the answer to that is 491 which is just below 500 and rest of the options will give above 500.
So, the 24th term will be 491 the last term which is less than 500 and also satisfies the condition.
1) The first minimum common number which will satisfy the condition is 8.
2) After that the second largest will be 29(8 + LCM of 7 and 3).
3) Now according to the condition the integers are less than 500 and the numbers after 29 will be in a A.P, so I used an options and used the A.P formula.
4) So, the option C 24, in the formula will be a21 = 8+23(21), common difference is 21 and the answer to that is 491 which is just below 500 and rest of the options will give above 500.
So, the 24th term will be 491 the last term which is less than 500 and also satisfies the condition.
KarishmaB
