Mind Your 5's And 7's

It is given that exactly 5 nos are divisible by 5 and exactly 7nos are divisible by 7. So, there are two nos which are divisible by both 5 and 7. The least possible of such nos are 35 and 70. Hence the value of M is 70

Anyone who is thinking that there are infinitely many solutions to this think again the problem is asking for the MINIMUM possible value of the greatest number on the list and as others have explained we have to have 2 numbers divisible by both 5 and 7 that is divisible by 35 so we have to have 35 and 70 on the list.We could have had 70 and 105 also but the question is asking the minimum value of the greatest number.The rest of the numbers that is the 3 numbers that have to divisible by only 5 and the 5 numbers that have to be divisible by only 7 can be choosen from numbers smaller than 70. To understand this better you need to think in terms of sets i.e N(A union B)=N(A)+N(B)-N(A intersection B)

There are 5 numbers that are divisible by 5, and 7 numbers divisible by 7. If these were all distinct (i.e. no number is divisible by both), there could not be only 10 numbers, since 5 + 7 = 12 numbers. Thus, at least two numbers are divisible by both 5 and 7.

The first of these is 35. The second is 70. 70 is the least number we know has to be included in the 10 numbers, so we can set M = 70.

it is specified that there are 5num divisible by 5 and 7 numbers divisible by 7 so num may be 5,10,15,7 ,14,21,28,42,35,70

The required set of Ammar may be: {5,7,10,14,15,21,28,35,42,70} Here the elements {5,10,15,35,70} are exactly 5 integers that are divisible by 5 and the elements {7,14,21,28,35,42,70} , two elements must be common multiples of 5 and 7 so here {35,70} are common multiples.

very simple. total numbers=10 5=5 and *7=7. so 2 numbers r common. common numbers =35(5 7) and 70(5 7 2) [least 2 common multiples] in remaining 8 numbers,4 r multiples of 5 and 4 r of 7. the question is asked such that M is min.so the answer is 70 :)

Find LCM = 35.....but there is no 35....so x 2...70...which can be divided by 5 n 7

10 distinct positive integers!!! at least 2 numbers divisible with both 5 and 7..... these two would be ....0 , 70

Total 10 integers are there hence two integers should be multiples of both 5 and 7.. That is 35 and 70.. Greatest of them is 70. Other 8 integers could be 5,10,15,7 ,14,21,28,42.

it is given 5 nos are divisible by 5 and 7 nos are divisible by 7 so there is minimum 2 nos divisible by both 5 and 7 . so least value for that is 35 and 70. thus minimum value of largest no. is 70.