Growing digits

Logic Level 1

$\begin{array} { rrrrr } & & & & & A \\ & & & & A & A \\ & & & A & A & A \\ + & & A & A & A & A \\ \hline & \large\square & \large\square & \large\square & \large\square & \large\square \\ \end{array}$

The above is an incomplete cryptogram, where $A$ is a positive, single-digit integer. As shown, the final sum is a 5-digit number.

What is this final sum?

The answer is 11106.

16 solutions

Michael Coley
May 11, 2017

By observation, the answer is 1234A. Since the answer is 5 digits (10,000 - 99,999), A must be between 8.1 and 81.03 but since we've been told it's a positive single-digit integer, the only valid value for A is 9. $9 \times 1234 = 11106$

For my education, am I correct that you have to assume base 10 to arrive at a solution? I haven't had any math since high school which was a long time ago. It's really great to benefit from the knowledge of those more mathematically literate than me.

Leroy Plock - 4 years ago

Yes, we assume numbers are written in decimal (base 10), unless stated otherwise.

Pi Han Goh - 4 years ago

4 * 9 = 36 ie first digit of the five digit number is 6; 3 * 9 +3 - 30 second digit is 0; 2*9+3 = 21; third digit is 1; 2 * 9 +2 = 20 so I got 20106 as the five digit number

Soundara Rajan - 4 years ago

Please check your calculation again. For the fourth digit, it should be 1 * 9 + 2 = 11, so the number is 11106

Pranshu Gaba - 4 years ago

tot = {}
x = 0
for i in range(1,10):
    for j in range(1,5):
        x += int(str(i)*j)
    tot[i] = x
    x = 0
print tot

1	`{1: 1234, 2: 2468, 3: 3702, 4: 4936, 5: 6170, 6: 7404, 7: 8638, 8: 9872, 9: 11106}`

Michael Fitzgerald - 3 years, 1 month ago

Azadali Jivani
May 9, 2017

Relevant wiki: Cryptogram

4A + 30A + 200A + 1000A = 1234A
put A = 9
therefore...... 1234 * 9 = 11106

Why must A=9? why can't it be 1,2,3,4,5,6,7 or 8?

Pi Han Goh - 4 years, 1 month ago

Because then the result would not be a 5-digit number.

Siva Budaraju - 4 years, 1 month ago

Can you prove that without trial and error?

Pi Han Goh - 4 years, 1 month ago

@Pi Han Goh – yes; all you have to test is 1234*8=9872,so A≠8, and if A<8, then obviously the result can't be any greater than A=8.

Siva Budaraju - 4 years, 1 month ago

@Siva Budaraju – Thank you very much. ♥

Melissa Lima - 4 years, 1 month ago

@azadali jivani I didn't understand why you did "4A + 30A + 200A + 1000A = 1234A". Why those number multiplying A?

Melissa Lima - 4 years, 1 month ago

There are 4 A's in the ones place, 3 A's in the tens place, 2 A's in the hundreds place, and 1 A in the thousands place.

Siva Budaraju - 4 years, 1 month ago

Thank you!

Melissa Lima - 4 years, 1 month ago

Why the numbers multiply by 9 why not do adding

Mansi Joshi - 4 years ago

What do you mean?

Siva Budaraju - 4 years ago

A Former Brilliant Member
May 14, 2017

The trick here is to set $A$ equal to the highest single digit which is $9$ . Then observe the result. If the final sum is 6-digit number, then set $A$ equal to $8$ , until you find a sum that is a 5-digit number. But setting $A$ equal to $9$ gives a sum of a 5-digit number. Therefore, the answer is $11106$ .

Whts the answer

Kamal Malhotra - 4 years ago

Really very poorly written

Kamal Malhotra - 4 years ago

Care to elaborate?

Pi Han Goh - 4 years ago

You have only shown that the answer is true when A=9 and when A≠8. Why can't A=1,2,3,4,...7 be the answer as well?

Pi Han Goh - 4 years ago

You don't get 5 digit numbers as a sum from 1-7

Chris Fistonić - 4 years ago

Very poorly written.

Kelly Waters - 4 years ago

Esperandieu Cetoute
May 20, 2017

To gain a final sum that contains 5 digit numbers, the only value to give to A must be 9. Try other one, you won't find a final sum which would have 5 digit numbers! See by yourself!!!!!!

Yes, but is there a faster approach without trying out A =1,2,3...,8 first?

Pi Han Goh - 4 years ago

Steve Peters
May 19, 2017

by allowing A to equal any single digit 1 through 8, the sum does not equate to a 5-digit number. Therefore, a must equal 9, and as a result

9+99+999+9999=11106

Right. Simple. The only single digit number that is 10 or greater after adding one.

Luke Kingson - 4 years ago

Samuel Qin
May 19, 2017

The first digit of the answer must be a 1. In addition, the carry over of the hundreds digit can be at most a one. However, for there to even be a number in the ten-thousand place, the hundreds must carry a one. Thus, the "A" must be 9. Therefore, the answer must be 9+99+999+9999=11106

Actually, the carry over of the hundreds is 2 in this case, yielding 9+2 = 11 as thousands and ten thousands digit! So your reasoning is off at that step.While adding more than two numbers at once, carry over can be greater than 1!

Martin Ramsch - 3 years, 4 months ago

Mohammad Khaza
May 16, 2017

if the answer is a five digit number then 8 is incorrect cz it gives a four digit answer.so,1 to 8 is invalid .then obviously 9 is the digit.and the answer is 9+99+999+9999=11106

This is a correct approach.

Is there a faster way to solve this without trying out all 9 integers?

Pi Han Goh - 4 years ago

ha ha.i did not tried from 1 to 8.at first i took 8. then i understood 1to 7 is invalid.then i took 9.if you know another way comment to me i want to learn a lot

Mohammad Khaza - 4 years ago

then i understood 1to 7 is invalid

Why must this be true? Just because 8 cannot be true, why 1 to 7 cannot be true either?

Pi Han Goh - 4 years ago

@Pi Han Goh – bro,if a bigger value cannot give wanted answer.then how summation of smaller values can give ?

Mohammad Khaza - 4 years ago

nice explanation

Halima Tahmina - 4 years ago

thanks.your comment inspired me

Mohammad Khaza - 4 years ago

Hetansh Mehta
May 15, 2017

the final sum can only be a 5 digit number,given A is a positive single digit if and only if A=9,substituting we get value of the sum =11106

Why must A=9? Why can't it be something else?

Pi Han Goh - 4 years ago

Abidur Rahman
May 15, 2017

Intuitively, the highest single digit would be the most likely value for A. This means that you would have 9999 + 999 + 99 + 9 which is equal to (10,000 + 1000 + 100 + 10) - 4. This is equal to 11106.

Your intuition is correct. Can you prove that it's indeed correct?

Pi Han Goh - 4 years ago

From right to left, you have 4A + 3(10A) + 2(100)A + 1(1000)A and this gives you 1234A. Given that 1234A is a five digit number, 1234A>=10,000, Therefore A>= (10,000/1234). In other words, A>= 8.1037.... Therefore A=9 as it is the only integer that satisfies these conditions.

Abidur Rahman - 4 years ago

Wonderful work!

Can you generalize this question? That is, what happens if there are $n$ rows of "increasing number of A's", and the final sum is an $n+1$ -digit number?

Pi Han Goh - 4 years ago

@Pi Han Goh – I may not have the required knowledge to solve this problem as I'm 16 and I really shouldn't be investigating the problem when I have my exams to focus on, so ... if you can show me how to solve it that'll be cool !

Abidur Rahman - 4 years ago

@Abidur Rahman – Don't worry. Everyone learns at their own page. Think about this in a couple of days first, if you're still stuck, come back to me.

Good luck with your exams...

Pi Han Goh - 4 years ago

@Pi Han Goh – I don't think the problem is the rate at which I'm learning, just that I'm still at the age where I don't really need to learn this in the first place. I try to solve these problems for fun and if I really try I'll get it eventually.

Here, let me give you a problem, (100!)x(99!)x(98!)x(97!) ... (3!)x(2!)x(1!), which factorial should you cross out, so that the result is a perfect square.

Abidur Rahman - 4 years ago

@Abidur Rahman – It's not about whether you need to learn this, it's about whether you enjoy learning them. It's no point learning random stuffs if you don't enjoy doing it.

For your factorials question: Start small, maybe 1! x2! ... x 8!, (instead of up to 100!), try to answer this simpler question first. Then spot a pattern. Then, the next step (and the fun step) is can you prove it?

Pi Han Goh - 4 years ago

@Pi Han Goh – Actually I already have an answer for this question, I'm asking you how you would solve it and then I will see if we got the same answer and also I do enjoy it, but learning and finding it out for yourself is different. I try not to learn the methods to solve maths, but instead find my own way around the problem or try to figure out what I'm supposed to learn before someone teaches it to me. That way I know that I can solve problems without needing to gain the knowledge beforehand.

Abidur Rahman - 4 years ago

@Abidur Rahman – If you would like to see how others solve your question, go ahead and post it as a problem instead.

Pi Han Goh - 4 years ago

@Pi Han Goh – See that doesn't work in this website unless I am certain about the answer. That's why I'm appealing to people in the comments for some certainty.

Abidur Rahman - 4 years ago

Venkatachalam J
May 15, 2017

Relevant wiki: Cryptogram - Problem Solving

It is easy to solve by converting the given Addition problem to equivalent Multiplication problem.

The only possible single digit positive integer value of A=9

How do you know that this is true? Did you do trial and error for the numbers 1,2,3,4,...,9?

Pi Han Goh - 4 years ago

Tried the maximum value "9" first and found the solution is working. For "8" it will give 4 digit result. So the only possible single digit positive integer value of A=9.

Venkatachalam J - 4 years ago

So you have shown that A=9 works and A=8 doesn't. But what about A=1,2,3,4,..,7? Why did you skip these are other numbers?

Pi Han Goh - 4 years ago

@Pi Han Goh – A=8 itself have only 4 digit. So no need to check other digits. Definitely 1,2,3,...8 will not give a five digit number. In the result we are expecting an 5 digit number.

Venkatachalam J - 4 years ago

@Venkatachalam J – No, that's not the right explanation. You need to show that f(A) = 1234A is an increasing function, so if f(8) < 10^5, then so does f(1), f(2), ... f(7) as well.

Pi Han Goh - 4 years ago

@Pi Han Goh – In my solution I clearly mentioned "The only possible single digit positive integer value of A=9" satisfy the condition. That means all other numbers are not fulfilling the required condition.

Venkatachalam J - 4 years ago

@Venkatachalam J – But you didn't explain why that's the case. It's like saying "The answer must be this, believe me."

Pi Han Goh - 4 years ago

@Pi Han Goh – It is understandable. If there are multiple solutions maybe we need to say the boundary conditions.

Venkatachalam J - 4 years ago

@Venkatachalam J – No, when you're writing a solution, you need to show that your solution is complete, not "Oh, because Brilliant only accept one particular value, then the answer must be unique."

Pi Han Goh - 4 years ago

@Pi Han Goh – My solution is complete. Are you find any other values satisfying the required condition. The equation only satisfy when we substitute A=9.

Venkatachalam J - 4 years ago

@Venkatachalam J – Like I said, you have only shown that A=9 works. But you didn't show that A=1,2,3...8 doesn't work.

Your solution could easily be fixed if you said "by trial and error, by trying out A=1,2,3,4...,9, only A=9 works"

Pi Han Goh - 4 years ago

@Pi Han Goh – It imply the meaning it will not work for other numbers.

Venkatachalam J - 4 years ago

Eric Lee
Feb 11, 2020

n = [i for i in range(1,10)]
def cryptogram(n,step):
    sub,res = 0,[]
    for x in n:
        sub = 0
        for i in range(1, step+1):
            print(sub)
            sub += int(i*str(x))
            re = (x,sub)
        res.append(re)
    return res

Rocco Dalto
May 17, 2017

Adding we obtain:

$S = [\displaystyle\sum_{j = 0}^{3} 10^{j} + \displaystyle \sum_{j = 0}^{2} 10^{j} + \displaystyle \sum_{j = 0}^{1} 10^{j} + 1] * A = [\dfrac{10^{4} - 1}{9} + \dfrac{10^3 - 1}{9} + \dfrac{10^2 - 1}{9} + 1] * A$

$= (1111+ 111 + 11 + 1) * A = 1234 * A.$

For $(1 <= A <= 8)$ , the sum $S$ is a four digit number. For $A = 9$ the sum $S = 1234 * A = \boxed{11106}$ .

Is there a faster way to show that the only solution for A is A=9 without trial and error?

Pi Han Goh - 4 years ago

Oli Hohman
May 16, 2017

In order for the sum of a 4 digit number and 3 digit number to equal a 5 digit number where each digit is the same, every digit must be a 9 (if each digit were 8, the sum would be 4 digits). Thus, the answer is 9+99+999+9,999 = 11,106.

Right, if $A$ is 8 or less than 8, the sum is a 4 digit number.

Pranshu Gaba - 4 years ago

Hjalmar Orellana Soto
May 16, 2017

The answer is $1234A$ and it needs $A+1=??$ (two-digits number), so $A=9\Rightarrow 1234A=11106$

I don't understand.

What does (A+1=?? two digits number) mean?

Pi Han Goh - 4 years ago

I meant $A+1$ equals a two digits number, otherwise the aswer shouldn't have 5 digits

Hjalmar Orellana Soto - 4 years ago

Why must A+1 equals a 2 digit number?

Pi Han Goh - 4 years ago

@Pi Han Goh – Whoa, really don't know how to explain it in english.... in spanish we use the word "decena" for the digit $a$ such that a number can be written as $x_{0}+10a+100\cdot x_{0}+\cdots$ .... the point is that when we sum two numbers as above and the sum of two digits is a two digit number (the greatest possibility) the "decena" is summed up with the following numbers.... don't know if i'm expressing as well xD anyways, is a solution more suitable for "number theory" than for "logic"

Hjalmar Orellana Soto - 4 years ago

@Hjalmar Orellana Soto – Yup, I know what you're trying to say.

It's called "carry over."

Pi Han Goh - 4 years ago

@Pi Han Goh – ok, thank you n.n

Hjalmar Orellana Soto - 4 years ago

Ishwar Karthik
May 15, 2017

The answer is $4A + 30A + 200A + 1000A$ or $1234*A$ . A 5 digit # has to be above 9,999, and $9999/1234 = 8.102...$ So A has to be 9. $9*1234 = 11106$ . QED

Nicely done.

Can you generalize this question? That is, what happens if there are $n$ rows of "increasing number of A's", and the final sum is an $n+1$ -digit number?

Pi Han Goh - 4 years ago

Noel Lo
May 15, 2017

Nice problem! Love it!! +1

Thanks Glad you liked it! :-)

Pi Han Goh - 4 years ago

Growing digits

The answer is 11106.

16 solutions

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