I want the answers to these 2 mathematical riddles.please help me?

Question

1.The sum of four consecutive two-digit odd numbers, when divided by 10, becomes a perfect square.Which of the following can possibly be one of these four numbers ?
a.21 b.25 c.67 d.73


2.The number of solutions of the equation 2x+y=40 where both x and y are positive integers and x=or < y is :
a.7 b.13 c.18 d.20


Please help me, with the derivation too, I am not able to solve it, I tried so much and need help.
Thank you vey much.

Answer 1

>1.The sum of four consecutive two-digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers ?
a.21 b.25 c.67 d.73

Call the numbers x, x+2, x+4, x+6
then their sum divided by 10 is:
((x) + (x+2) + (x+4) + (x+6))/10 = (4(x+3))/10 = 2(x+3)/5
and we are to test if that is a square.
So:
- if the number we are testing is x, we test 2(x+3)/5
- if the number we are testing is x+2, we test 2(x+1)/5
- if the number we are testing is x+4, we test 2(x-1)/5
- if the number we are testing is x+6, we test 2(x-3)/5

A weaker test is simply for it to be an integer, so if (x±1)or(x±3) doesn't have a factor of 5, 2(x+3)/5 isn't even an integer.
=> this implies the factor of 5 must come from the (x±1)or(x±3). So only consider the brackets which are divisible by 5.
=> immediately we can eliminate 25

Plugging x=21 into -> 2(x-1)/5 gives 8, not square
Plugging x=73 into -> 2(x-3)/5 gives 24, not square

ANSWER: none

2.The number of solutions of the equation 2x+y=40 where both x and y are positive integers and x=or < y is :
a.7 b.13 c.18 d.20

0 So set y = x+h
Then: 2x+y = 3x+h = 40
Then there are 13 ways we can pick x=1..13 such that
x>0 and h is still positive

ANSWER: b

Answer 2

2.The number of solutions of the equation 2x+y=40 where both x and y are positive integers and x=or < y.
To solve this rewrite the equation
y=40-2x
and plug in different values for x (starting at x=1), see what you get for y, until you reach a value where y So possible values for x:
1,2,3,4,5,6,7,8,9,10,11,12,13,14 -wait we've gone too far - if x=14, then y =12
So there are 13 possible solutions.

1.The sum of four consecutive two-digit odd numbers, when divided by 10, becomes a perfect square.Which of the following can possibly be one of these four numbers ?
a.21 b.25 c.67 d.73

Let the numbers be x-2,x,x+2,x+4
Sum = 4x+4
sum/10 = perfect square
sum = perfect square x 10
sum = 40, 90, 160,250, 360 we can't go past here because they are only 2 digit numbers - and we can't actually have 40 either because of the same reason.
So solve the equation 4 times.
4x+4 = 90 , 160, 250, 360
4x = 86, 156, 246,356
These aren't all divisible by 4
The only possibilities are 156 and 356
so x = 39 in which case the consecutive odd integers are 37, 39, 41 and 43
or x= 89 in which case we have 87, 89, 91,93
Hey looks like none of your options work.

Answer 3

Well, here you go, but you're not gonna like it. The answer to the first, is none of the above. The first Number when divisible by 10 which is a perfect square is 160, and that is made up of 4 consecutive odd numbers 37+39+41+43. the next and only possible choice ( 2 digit odd numbers limitation ) is 360, made up of 87+89+91+93. Hence, none of the answers listed is correct. But as for the second, the correct answer is 13. This is because, since x<= y then sum of 2x + y where x<=y, then 2x + x must be <= 40, or 3x<= 40, hence x <=13. therefore x has a range of ( 1 to 13 ), and 13 possible solutions...

Answer 4

1.When divided by 10, the sum of 4 consecutive 2 digited odd numbers should be a perfect square.

So, the perfect square Should be a 1 digit or a 2 digit number.(think)

So, the sum should be either of 90,160,250,360.
(squares of numbers multiplied by 10 such that sum is below 400).

So, check it with your answers. Multiply your options with 4, and add or subtract 12 or 4(WHY? Think. simple) and verify if anything of these matches.

I feel none of them are matching.

87,89,91,93 form one quadruplet.(mean=90=360/4)
37,39,41,43 form another.(mean=40=160/4)

And nothing else.

2. X is Then solve the equation. u have 3x=40.
so, x can be 13. and y=14.(only integers).
So, the no. of solutions of the equation satisfying the above criterion will be 13.

(1,38) (2,36) (3,34) (4,32) (5,30) (6,28) (7,26) (8,24) (9,22) (10,20) (11,18) (12,16) (13,14).

Answer 5

((2n + 1) + (2n + 3) + (2n + 5) + (2n + 7))/10 = m^2
(8n + 16)/10 = m^2
4(n + 2) = 5m^2
n = 5(m/2)^2 - 2
2n + 1 = (5/2)m^2 - 3
Because of the division by 2, m must be even.
m, numbers
2 . . 7 . . 9 . 11 . 13
4 . 37 . 39 . 41 . 43
6 . 87 . 89 . 91 . 93
8 157 159 161 163
None of the choices given work.

2.
2x + y = 40
x ≤ y
y = 40 - 2x
x ≤ 40 - 2x
3x ≤ 40
x ≤ 13
so the number of solutions is 13