As small as possible

Number Theory Level 5

Four positive integers $a,b,c,d$ satisfy $abcd=10!$ .

What is the smallest possible value of $a+b+c+d ?$

Notation: $!$ is the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$ .

The answer is 175.

2 solutions

Brian Charlesworth
Mar 19, 2017

First note that $10! = 2^{8} \times 3^{4} \times 5^{2} \times 7$ . To find the least possible sum of 4 factors of $10!$ , we are looking for the 4 factors closest to the 4th root of $10!$ , which is approximately $43.65$ . Apportioning the prime factors optimally, the 4 factors closest to this value are

$2^{3} \times 5 = 40, 2 \times 3 \times 7 = 42, 3^{2} \times 5 = 45$ and $2^{4} \times 3 = 48$ .

Thus $a + b + c + d = 40 + 42 + 45 + 48 = \boxed{175}$ .

yes sir that is correct and your initial solution was also detailed and more informative as it also provided the four numbers when the least possible sum will take place. thank you sir

Sathvik Acharya - 4 years, 2 months ago

Hows it by soing AM GM sir?

Md Zuhair - 4 years, 2 months ago

Yes, good point. As Sathvik is also implying, we know that

$a + b + c + d \ge 4 \times (10!)^{1/4} \approx 174.58$ .

Since $a,b,c,d$ are integers this then means that $a + b + c + d \ge 175$ ,

and the minimum can be achieved when $(a,b,c,d) = (40,42,45,48)$ .

Brian Charlesworth - 4 years, 2 months ago

It is very clear that 1 can go with any of a, b, c, or d. So we will not consider 1.
It is clear that say group G1={10, 9, 8, 7} in {a,b,c,d} in same order to achieve minimum
by putting the four biggest numbers in separate letters.
Now the next group G2={ 3,4,5,6} in { a,b,c,d} in same order to achieve minimum product between elements of same letter.
Here is the catch. Here six can be obtained in TWO ways. 2 * 3 and 6 alone. (I missed this point!!)
Naturally, any one will be with 7 and the other with 8. So that now, corrected G2={4,5,6, 2 * 3}.
So G1 + G2={10 * 4, 9 * 5, 8 * 6, 7 * 3 * 2}= {a,b,c,d} and 1 with any one of the letters.
So a+b+c+d = 40 + 45 + 48 + 42 = $\color{#D61F06}{175}.$

Niranjan Khanderia - 4 years, 2 months ago

yeah that's a good solution but you can even find the answer without finding a,b,c and d by just approximating 4 times the 4th root of 10!.

Sathvik Acharya - 4 years, 2 months ago

Md Zuhair
Mar 21, 2017

The Minimum can be achieved Like Brian Sir, Did it, And I suggested this meathod, I did it like this

By AM - GM Inequality

$\dfrac{a+b+c+d}{4} \geq (10!)^{\frac{1}{4}}$ Making It

$a+b+c+d \geq 4 \times (10!)^{\frac{1}{4}}$

So we get $\text{Minimum of a+b+c+d} \approx 174.58$

So as a , b ,c ,d are integers

So $Min(a+b+c+d) = 175$ and it is minimum at $(a,b,c,d) = (40,42,45,48)$ .

sir can you please tell me how do we arrive at the values of a,b,c and d. is it that we have to make the factors as close as possible but my doubt is can we find a,b,c and d in one shot without checking whether there are other numbers that give a better minimum sum

Sathvik Acharya - 4 years, 2 months ago

Its not possible I think. I just got it from @Brian Charlesworth sir. Also dont call me sir. I am a Xth standrd bou

Md Zuhair - 4 years, 2 months ago

I believe anybody who has a greater knowledge than himself/herself deserves to be called with respect. thank you

Sathvik Acharya - 4 years, 2 months ago

Once the AM-GM analysis indicates that the minimum possible sum is $175$ , and then we are able to find $a,b,c,d$ such that $a + b + c + d = 175$ , we don't have to worry about there being numbers that give a better minimum sum. As for finding the specific optimal numbers for $a,b,c,d$ , that just took a bit of juggling with the prime factors of $10!$ .

Brian Charlesworth - 4 years, 2 months ago

As small as possible

The answer is 175.

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