Just A

Logic Level 1

A A A A A A + A 1 0 0 \begin{array} { l l l l } & A & A \\ & A & A \\ & & A \\ & & A \\ + & & A \\ \hline 1 & 0 & 0 \end{array}

What is A A ?

2 4 6 8 0

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Chung Kevin by Chung Kevin , 45, Australia

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15 solutions

Venkatachalam J
Mar 27, 2017

43 Helpful 0 Interesting 0 Brilliant 0 Confused

why 10? in the beginning the only thing in my mind was A times A = A square so 2(a)^2 + 3 a = 100 but got confused a is 10 a how did you find that

Joanna Demee - 4 years, 2 months ago

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Why 10 in the beginning?

Alshaimaa Elgenbihy - 4 years, 2 months ago

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Because it is in the tens place. AA can be written as = (A x10) +A =10A + A in the expanded number form using base 10. – Venkatachalam J

Venkatachalam J - 4 years, 2 months ago

It is in the number expanded form. "A" is a single digit. In "AA" the left most "A" is in tens place and right most is in the once place. Therefore it is written as "10A+A". If you check with any two digit number you will get it quickly. For example the number "27" can be written in the expanded form (2x10)+7=20+7.

Venkatachalam J - 4 years, 2 months ago

Nice. I had once seen a similar cryptogram where there were 8 8's that made a 1000.

Agnishom Chattopadhyay - 4 years, 2 months ago

Based on the comments I updated the required modification. I hope, it will clear and understandable.

Venkatachalam J - 4 years, 2 months ago
Zee Ell
Mar 22, 2017

( 10 A + A ) + ( 10 A + A ) + A + A + A = 25 A (10A+A) +(10A+A) + A+A + A = 25A

25 A = 100 25A = 100

A = 4 A = \boxed {4}

40 Helpful 0 Interesting 0 Brilliant 0 Confused

25 plus 4 equals 100

Prabhbir sohi - 4 years, 2 months ago

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😂 25 + 4 = 29

25 × 4 = 100

Zee Ell - 4 years, 2 months ago
Ahmed Amir
Mar 27, 2017

5A + 20A = 100

A = 4

10 Helpful 0 Interesting 0 Brilliant 0 Confused

@Agnishom Chattopadhyay You can represent numbers like 55, 31, 72, etc... as 5 * 10+5, 3 * 10+1 and 7 * 10+2 respectively. Try and think how that fact applies to this problem.

Ebuka Akubilo - 4 years, 2 months ago

Could you explain how you got the 5 and 20 in 5A + 20A?

Agnishom Chattopadhyay - 4 years, 2 months ago

To a novice, there is nothing obvious about this solution. No idea what the different columns represent, how to interpret them, or how they interact. Back to 2+2=4, my favorite.

Carson Foard - 4 years, 2 months ago
Mervyn Rice
Mar 29, 2017

44+44+4+4+4= 100

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Yes you have shown that A=4 is a solution, but is it possible that there are other possible solutions as well? Why or why not?

Pi Han Goh - 4 years, 2 months ago
Dominic Dharam
Apr 2, 2017

3A + 11A +11A =100 A (3+11+11) =100 A = 100/25 = 4

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Galen Buhain
Apr 2, 2017

5(a) + 20(a) = 100 a = 4

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Guy-Philippe Goth
Mar 29, 2017

11xA+11xA+3xA=100

25A=100

A=4

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Guillermo Zamudio
Mar 28, 2017

This is assuming that you have a base 10 system. if the base was0, the A could also be zero. which is a silly problem ut points out that a second solution is real, depending on the base used. A=b *2/(2 b+5) b being your base.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

I like your thinking :)

If the base is a positive integer, what other solutions can we have (where A is also an integer)?

Chung Kevin - 4 years, 2 months ago
Avianna Gay
Apr 2, 2017

Just by glancing at it, we know that 0 and 2 are automatically out , because the numbers they yield are TOO SMALL:

0 ----> (00+00=0)

2----> (22+22+2+2+2=44+6<100)

In the same instance, we also know that that 6 and 8 are automatically out due to the fact that the numbers that they yield, are too LARGE to begin with. Doubling the double digit numbers, alone, would yield a number GREATER than 100:

6 ---> (66+66>100)

8 ---> (88+88>100).

Therefore, the only logical answer must be 4:

4 ---> 44+44+4+4+4=88+12= 100

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Dibakar Sunny
Apr 2, 2017

0 Helpful 0 Interesting 0 Brilliant 0 Confused
冠璋 陳
Apr 1, 2017

Why would 11% get this wrong?

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Tarick Hill
Mar 31, 2017

i got the right answer but i did it a completely different way i took it as an a and another a and since they were right next to each other i took the 2 a's and put it as 44 and 44 times 2 is 88 and 88 + 12 = 100

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Sorry, I did not understand what you meant.

Agnishom Chattopadhyay - 4 years, 2 months ago
Gabby Dackers
Mar 30, 2017

A=4 . 6 and 8 are too high a value/s, and 0 just equals 0. This leaves 2 and 4. However, A cannot be 2 because 22+22+2+2+2 will only equal 50. So it makes sense that when you double the value 2 to make 4, you double the result of the AA + AA + A + A + A, and so 50 becomes 100. Also. 2 lots of 44 (= to AA) equal 88, and 3 lots of 4 (= to A) equal 12. 88 + 12 = 100

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Yeah, so out of all the options, A = 2 works, but can A be equal to 1,3,5,7, or 9 as well?

Pi Han Goh - 4 years, 2 months ago
Cassidy McGuire
Mar 29, 2017

I just went through each number. A = 2 isn't it as it would equal fifty A = 6 or 8 aren't it, as they would equal numbers over 100. A = 0 would just be 0 So A = 4

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Yup, A=4 is correct. But is there a more systematic way to solve this without resorting to trial and error?

Pi Han Goh - 4 years, 2 months ago
Emily Lessmiller
Mar 28, 2017

just find the larger two and the last ones become easier. try all the numbers displayed to help. if AA=2, then it would total to 50, not 100. if you took 6, then you would find that wouldnt work, right away. 66+66=132, and thats over 100 already. plus, 0 added to itself makes 0, so thats not the answer. the answer is, AA=44, so 44+44= 88, and A+A+A= 12, (4+4+4=12) and 88+12=100. Its not that complicated. Isnt this supposed to be hard? I could have solved this in sixth grade, and right now I'm in eighth. Take that for a smart kid.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Yup, you have experimented for all the possible values of A, but is it possible to get the answer without experimenting with all these options?

Pi Han Goh - 4 years, 2 months ago

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