\(3AA1⋮9\) => \(3+A+A+1⋮9\) => \(4+A.2⋮9\) => \(4+A.2=\left\{9;18;...\right\}\)

If \(A=1\) then \(4+A.2=4+1.2=6\left(wrong\right)\)

...................

We will have \(A=7\)

From the question, we have:

\(a:54=c\left(r=38\right)\) and \(a:18=14\left(r\ge0\right)\)

We see from the question: \(14.18\le a\le14.18+17\) <=> \(252\le a\le269\)

If \(c=1\) in the first equation => \(a=54+38=92\left(wrong\right)\)

If \(c=2\) in the first equation => \(a=54.2+38=146\left(wrong\right)\)

............

If \(c=4\) in the first equation => \(a=54.4+38=254\left(right\right)\)

So \(a=254\) => \(r=2\).

So the answer of the question is: \(254+2=256\)

Your answer must be changed into: \(1\le x\le1000\) and the result is: \(\dfrac{\left(1000-1\right)+1}{2}=500\) .

I think that the positive integers \(\left(Z^+\right)\) is from 1 to more, CTC. 0 is not a positive integer.

We see: The number : The pairs = 2.

So there are: \(1000:2=500\left(pairs\right)\)

So the answer is: \(500\left(pairs\right)\)

All my question is the "NO CALCULATOR" question. (Except for the question wrote: Calculator accept)

I had fix it. Thank you for your comment.

This is the orignal question in my study book.

\(27.8.5^3.64=3^3.2^3.5^3.4^3=\left(3.2.5.4\right)^3=120^3\)

There are 2 types of the isosceles triangle:

+ Acute triangle: All base ankles are \(< 90degrees\)

+ Right triangle: There is a right base ankles \(\left(=90degrees\right)\)

However, an isosceles triangle must be smaller than 180 degrees, the total of all ankles.

The cost of 10 erasers and 5 pens are:

\(\left(1,75.5\right)+\left(0,85.10\right)=8,75+8,5=17,25\)

So the answer is: \(17,25\)

P/s: I can't write the symbol of dollar.

\(2448:\left[119-\left(x-6\right)\right]=24\)

\(\left[119-\left(x-6\right)\right]=2448:24\)

\(\left[119-\left(x-6\right)\right]=102\)

\(\left(x-6\right)=119-102\)

\(x-6=17\)

\(x=17+6=23\)

\(2,4+\dfrac{3}{5}-\left(-0,7\right)+1,23\)

\(=2,4+0,6-\left(-0,7\right)+1,23\)

\(=3-\left(-0,7\right)+1,23=3,7+1,23=4,93\)

The numbers from 100 to 118 satisfy the question.

So there are: \(\left(118-100\right):1+1=19\left(numbers\right)\)

So the answer is: \(19numbers\)

Using the hundreds digit 4, we have 6 numbers: \(406;408;460;468;480;486\)

Using the hundreds digit 6, we have 6 numbers: \(604;608;640;648;680;684\)

Using the hundreds digit 8, we have 6 numbers: \(804;806;840;846;860;864\)

If the hundreds digit is 0, then the numbers are not the 3-digit-number.

So there are: \(6+6+6=18\left(numbers\right)\)

At the end of the summer, his height has increased: \(140.5\%=140:100.5=7\left(cm\right)\)

So his height at the end of the summer measured in cm is: \(140+7=147\left(cm\right)\)

So the answer is: \(147cm\)

Because \(BC=5cm\) => \(AD=5cm\)

The total area of triangle AMD and triangle BCN is:

\(\left(\dfrac{2.5}{2}\right).2=10\left(cm^2\right)\)

Then the area of trapezoid MNCD is:

\(60-10=50\left(cm^2\right)\)

So the answer is: \(50cm^2\)

Trung Nguyễn, do you copy the solution in the link?

You can go to this link for the answer: https://mks.mff.cuni.cz/kalva/short/soln/sh9023.html

The greatest distinct 4-digit number is: \(9876\). The least 3-digit number is: \(100\).

So their sum is: \(9876+100=9976\)