I did this in another way. 27*2+2=56 27*3-2=79 So we see here that sum of multiplying factors is 5 and 2 is common. Similarly, 23*1+4=27 23*4-4=88 Here again sum of multiplying factors is 5 but common number is 4. So, 16*3+8=56 16*2-8=24 Here also sum of multiplying factors is 5 and common number is 8.
I have a question exactly like this In first circle there is 1,8,7 and in second there is 2,4,56 and last circle have 3,100,? Can anyone have solution for this logic
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no ways i could solve this, it is really difficult Maths Puzzle.
I did this in another way.
27*2+2=56
27*3-2=79
So we see here that sum of multiplying factors is 5 and 2 is common.
Similarly,
23*1+4=27
23*4-4=88
Here again sum of multiplying factors is 5 but common number is 4.
So,
16*3+8=56
16*2-8=24
Here also sum of multiplying factors is 5 and common number is 8.
If the numbers are a,b,c then
a*m+n=b
a*(5-m)-n=c
Adding gives a*5=b+c. So it's not a coincidence that this gives the same result.
I have a question exactly like this
In first circle there is 1,8,7 and in second there is 2,4,56 and last circle have 3,100,?
Can anyone have solution for this logic
Really good , different from other math 🧩
👍👍👍👍
Where is logic boss... In first circle sum is 135 and in 2nd 115
Ok
No you can telling wrong 😢😮😊😊