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When the numbers 5, 7, 11 divide a multiple of 17, the remainders left 18 Jan 2021, 00:49
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When the numbers 5, 7, 11 divide a positive multiple of 17, the remainders left are respectively 4, 6 and 10. Which positive multiple of 17 gives the least number that satisfies the given condition?

A. 315th
B. 316th
C. 317th
D. 384th
E. 385th


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When the numbers 5, 7, 11 divide a multiple of 17, the remainders left are respectively 4, 6 and 10. Which multiple of 17 gives the least number

a. 384 b. 317 c. 385
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C
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When the numbers 5, 7, 11 divide a multiple of 17, the remainders left 18 Jan 2021, 05:43
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We can see that the difference between the divisor and remainder is constant = 1
5-4=1
7-6=1
11-10=1

So the least number will be 1 less than the LCM (5,7,11),
LCM(5,7,11) = 385, Least such number=385-1=384

Since the number has to be a multiple of 17, it will be of the form 385n-1 = 17k
The least such number is 5389 = 17*317
So Option C
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When the numbers 5, 7, 11 divide a multiple of 17, the remainders left 04 Jun 2021, 06:26
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n/5 leaves remainder 4
n/7 leaves remainder 6
n/11 leaves remainder 10.
If n would have been 1 more , then it would be perfectly divisible by 5,7 and 11.
So least number which is divisible by 5,7,11 = LCM(5,7,11) = (n+1)
So n+1 = 385.
n = 384
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When the numbers 5, 7, 11 divide a multiple of 17, the remainders left 29 Oct 2021, 08:31
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rocky620
We can see that the difference between the divisor and remainder is constant = 1
5-4=1
7-6=1
11-10=1

So the least number will be 1 less than the LCM (5,7,11),
LCM(5,7,11) = 385, Least such number=385-1=384

Since the number has to be a multiple of 17, it will be of the form 385n-1 = 17k
The least such number is 5389 = 17*317
So Option C
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Could you please ealborate the colored part a little bit
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When the numbers 5, 7, 11 divide a multiple of 17, the remainders left 29 Oct 2021, 08:49
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ShaikhMoice
rocky620
We can see that the difference between the divisor and remainder is constant = 1
5-4=1
7-6=1
11-10=1

So the least number will be 1 less than the LCM (5,7,11),
LCM(5,7,11) = 385, Least such number=385-1=384

Since the number has to be a multiple of 17, it will be of the form 385n-1 = 17k
The least such number is 5389 = 17*317
So Option C


Could you please ealborate the colored part a little bit
Show more

Check with options.
385 *n = 17k + 1

LHS will always end with 5 or 0. So k will end with 2 or 7 only. 317 is only such option

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When the numbers 5, 7, 11 divide a multiple of 17, the remainders left 29 Oct 2021, 09:26
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(1st) write out the Division Lemma equation for each piece of information

We are looking for a multiple or 17 (call it N for now) where:


N = 5a + 4

N = 7b + 6

N = 11c + 10

Where a, b, and c are non negative integer quotients


Each divisor gives the same exact Negative Remainder = -1

The first term that will satisfy each statement (ignoring the multiple of 17 requirement) will be -1 less than the LCM of the 3 divisors

N = LCM(5 ; 7 ; 11)* k - 1

Where k = non negative integer

N = 385k - 1


(2nd) we are given that N must be a multiple of 17 such that:

N = 17m

Where m = non negative integer

And

17m = 385k - 1


(3rd) at this point, since the answer choices are favorable, we can check one answer from (A) or (B) or (C) and then adjust based on how far from the condition we are

And then one answer from (D) or (E) if none of the first 3 satisfy the conditions

Or, we can use a Units Digit Analysis too see which answer choice must work

Since the question asks for the least multiple of 17, start with the first 3 answer choices.


1st positive multiple of 17 ——— 1 * 17 = 17

2nd positive multiple of 17 ——- 2 * 17 = 34

And so on

We are looking for the least positive multiple of 17 given:

17m = 385k - 1

385k = 17m + 1


On the LHS of the equation, the factor 385 multiplied by any non negative integer k will yield a Units Digit of either 0 or 5

Therefore, the RHS must yield a Units Digit of 0 or 5

In other words ——-> 17m + 1 must yield a Units Digit of of 0 or 5

(A)315th positive multiple of 17

(17) (315) + 1 ————> units digit of 6

(B)316th positive multiple of 17

(17) (316) + 1 ———-> units digit of 3

(C) 317th positive multiple of 17

(17) (317) + 1 ———-> units digit of 0

(C) is a possible answer!


(d) 384th positive multiple of 17

(17) (384) + 1 ———> units digit of 9

(e) 385th positive multiple of 17

(17) (385) + 1 ———-> units digit of 6


The only possible answer that could work is (c) 317th positive multiple of 17 = 5, 389


Condition 1:
5,389 ———-> divided by 5, remainder = 4

Condition 2:
5,389 ———-> divided by 7

4,900 is divisible by 7 ———> (5,389 - 4900) = 489

490 is divisible by 7, and 483 is divisible by 7

Thus the excess 489 when divided by 7 will yield a remainder = 6


Condition 3:
5,389 ————-> divided by 11

Remainder will be given by —-> (9 + 3) - (8 + 5) divided by 11

(12) - (13) = -1

-1 negative remainder is equivalent to a positive remainder of 10


Proof that all 3 conditions are satisfied


Answer (C)

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When the numbers 5, 7, 11 divide a multiple of 17, the remainders left 01 Nov 2021, 10:08
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Given : (17* Unkonwn)/5 => Reminder -->4

Multiply each no's last digit by 7 if 17

A. 315th --> 7*5 = 35 => 35/5 (Perfectly divide, no reminder)
B. 316th --> 7*6 = 42 => 30/5 (Perfectly divide, reminder --> 2)
C. 317th --> 7*7 = 49 => 49/5 (Perfectly divide, reminder --> 4) [Given, hence ans]
D. 384th --> 7*4 = 28 => 28/5 (Perfectly divide, reminder --> 3)
E. 385th --> 7*5 = 35 => 35/5 (Perfectly divide, no reminder)
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