P,Q < R
If (1+3+5+....+p) + (1+3+5+....+q) = (1+3+5+....+r) where each set of parenthesis contains the sum of consecutive odd integers as shown, and r > q > p > 6. Then the smallest possible value of p+q+r , is
The answer is 45.
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3 solutions
p = 11, q = 15, r = 19
(1+3+5+7+9+11) + (1+3+5+7+9+11+13+15) = (1+3+5+7+9+11+13+15+17+19)
100 = 100
p+q+r = 11+15+19 = 45
The answer is 45. how come the r will be smaller than p and q?
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yeah i m wrong i hav by mistake interchanged the value of q and r and thanx for bringing it to my notice
(1+3+5+⋯+p)+(1+3+5+⋯+q)=(1+3+5+⋯+r) (1+p)(((1+p))/2)+(1+q)(((1+q))/2)=(1+r)(((1+r))/2) (1+p)^2/2+(1+q)^2/2=(1+r)^2/2 (1+p)^2 + (1+q)^2=(1+r)^2
Since p, q and r are odd, then (1+p), (1+q) and (1+r) are even numbers that form a pythagorean triple. The smallest such triple is 12-16-20, so it follows that p=11, q=15 and r=19. Hence, p+q+r = 11+15+19 = 45
There are 2 n + 1 elements in the series 1 + 3 + 5 + . . . + n , so using the formula for the sum of the arithmetic series and a few algebraic transformations we get ( 1 + p ) 2 + ( 1 + q ) 2 + ( 1 + r ) 2 where ( 1 + p ) , ( 1 + q ) , ( 1 + r ) is a pythagorean triple. Since all three of these numbers are even, we look for a pythagorean triple of three even numbers a = 1 + p > 1 + 6 = 7 , b = 1 + q > 1 + 7 = 8 and c = 1 + r > 1 + 8 = 9 . It is a triple ( 1 2 , 1 6 , 2 0 ) , so p + q + r = ( 1 2 − 1 ) + ( 1 6 − 1 ) + ( 2 0 − 1 ) = 4 5
WE CAN SOLVE THIS AS :SUM OF CONSECUTIVE ODD NUMBERS STARTING WITH 1 ALWAYS RESULTS IN SQUARES AND BY PYTHOGORUS THEOREM THE SMALLEST PAIR IS 3,4,5 BUT HERE P WILL BE SMALLER THAN 6 SO WE FIND NEXT SMALLEST PAIR WHICH IS 6,8,10. THUS P=11,Q=15,R=19......CHEERS!!!!!
since p has to be greater than 6 the next smallest odd number is 7 so lets assume that p=7 . so we get on the RHS that r>7 . So the next greatest odd number is 9 so let us assume that r=9 . So thus by further solving we obtain that the lowest possible value of q will be 5 .
so thus,
p+q+r = 7+5+9 = 21