Fill in the blanks
Place the numbers 1 through 9 in the circles below, such that each side of the triangle adds up to 17.
Find those circles filled here!
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| Fill Numbers |
Find those circles filled here!
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Correct Numbers in Blanks
What was the challenge?
Just putting at 1,2 & 3 at 3 corners of triangle leaves only thing to find other 2 numbers giving total sum 17. Have a look at below.
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| Correct Numbers in Blanks |
The Alphamatic Problem
In the addition below, all digits have been replaced by
letters. Equal letters represent equal digits and different letters
represent different digits.
ABCABA
+
BBDCAA
+
ABEABB
+
ABDBAA
-------------------
AAFGBDH
What does the complete addition look like in digits?
Note : Alphamatic in the title is word derived from Alphabets & Mathematics. In such problems numbers are replaced by alphabets. The challenge is to find the number for each alphabet satisfying given mathematics equation.
Answer Of Alphamatic Problem
Here is question!
First of all let's write down the equation once again.
ABCABA
+
BBDCAA
+
ABEABB
+
ABDBAA
-------------------
AAFGBDH
We will refer to places in number from left as a first, second, third...sixth instead of tenth, hundredth, thousandth etc.
First we need to find if the 5 digits of first number itself i.e. ABCAB are carries forwarded from previous place.
From the addition of variable from first place, we get,
3A + B = 10A + A ........(1)
B = 8A ........(2)
Only numbers satisfying above are A = 1 & B = 8 , but at previous place we have addition of 4 B's. If B = 8, then addition at second place would be 32 with F = 2 & carry 3 which is not equal to A = 1. So A can't be a carry. So we need to modify (2) above as
B + x = 8A .........(2)....Where 'x' is carry forwarded from second place.
If B = 1 or 2 then x = 0 as at second place we would have F = 4 or 8. In that case, A would be fractional. Some other possible combinations for B, A & x are,
B = 9, x = 3, 8A = 12,
B = 8, x = 3, 8A = 10,
B = 7, x = 2, 8A = 9,
B = 6, x = 2, 8A = 8,
This is the only combination that can make A a whole number. So A = 1, B = 6.
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From sixth place, we have,
H = 3A + B = 9
---------------------------------------------------------------------------------------------
From fifth place, we have,
D = 2A + 2B = 14
But it has to be single digit i.e. D = 4 with 1 carry forwarded to next.
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From fourth place,
B = 2A + B + C + 1 .....1 is carry from last place.
6 = 9 + C
Now C can't be negative hence C + 9 has to be 2 digit number with 6 at last digit.Since addition of 2 single digit numbers never exceeds 18, C + 9 has to be 16.
16 = 9 + C
gives, C = 7 & carry 1 forwarded to third place.
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Unfair Loss in Savings of Life
On his 20th birthday, Ajay started depositing Rs.500 in his saving box. His brother Vijay used to withdraw Rs.100 from the same box on his birthday every year. On his 60th birthday, Ajay opens his box & finds only Rs.1000 in it.
How this could be possible?
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| Ajay & Vijay |
Having Birthday On Leap Day
Since Ajay celebrates his birthday on 29th February, he used to deposit Rs.500 in saving box once in 4 years. From age 20 to 60, he celebrated only 10 birthdays saving only 500 x 10 = 5000.
But Vijay withdrawing Rs.100 every year means he took 100 x 40 = 4000 so far.
That's why only 5000 - 4000 = 1000 left in the box.
This Is Out Of Box Thinking!
Why it was needed?
I said earlier this needs some out of box thinking.
If you carefully observe the digits in the series you won't find any digit exceeding 7 or 8. From that you have to conclude the series must be in octal (number system to base 7 is uncommon).
The series can be represented as,
(x + 1) ^2 ...........where x varies from 0,1,2,3,4,5,6,7,10........
And it's in octal number system.
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| Out of Box Thought |
So the answer is 241.
