Ans would be 40
I approached it this way
If there are only 7s then we need 50 of them.. but the answer revolves around 40 so there must be
at least one 77...
When there is one 77, remaining no is 350-77 = 273..Now check if this is factor of 7... yes it is
as 39*7 = 273
so total nos are 39 7s + 1 77 = 40
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Let's think about 7s and 0s.
We know that the sum ends in 0. We know that each term in the set ends in 7.
If we think about the multiples of 7, the ONLY time we end in a 0 is if we mutiply 7 by a multiple of 10. Therefore, the only way the sum of our set can end in a 0 is if we have 10, 20, 30, 40, 50, ... terms.
Only one answer matches our prediction of "ends in 0": choose (c) 40.
We know that the sum ends in 0. We know that each term in the set ends in 7.
If we think about the multiples of 7, the ONLY time we end in a 0 is if we mutiply 7 by a multiple of 10. Therefore, the only way the sum of our set can end in a 0 is if we have 10, 20, 30, 40, 50, ... terms.
Only one answer matches our prediction of "ends in 0": choose (c) 40.
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Another way to think about this Q
Sum = a1+a2...+an = 7x+77y (where x = number of 7 and y=number of 77)
= 7(x+11y) = 350
or x+11y = 50
y=1 then x=39 so T=40
y=2 then x=28 so T=30
y=3 then x=17 so T=20
y=4 then x=6 so T=10
Sum = a1+a2...+an = 7x+77y (where x = number of 7 and y=number of 77)
= 7(x+11y) = 350
or x+11y = 50
y=1 then x=39 so T=40
y=2 then x=28 so T=30
y=3 then x=17 so T=20
y=4 then x=6 so T=10
