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Sidana [21]
3 years ago
12

The sum of three consecutive integers is greater than 66. What is the smallest possible product of the largest and smallest of t

hese integers?
Please help me on this
Mathematics
1 answer:
AysviL [449]3 years ago
4 0

Answer:

528

Step-by-step explanation:

So naturally at first sight for this question we would think -->

Oh 3 consecutive integers = 66 --> 66/3 = 22 (n-1), (n+1) so --> 21, 22, and 23.

But no. At second look it is the sum of 3 consecutive integers is greater than 66. So we find the next possible pair since it says smallest possible product.

We get the set (22, 23, 24). => Multiply the least and greatest integers together respectively 22 and 24 which amounts to => 528

And thus, we have out answer of 528

Hope this helps!

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9.21 <br><img src="https://tex.z-dn.net/?f=9.21%20%5Cdiv%200.4%20%3D%20" id="TexFormula1" title="9.21 \div 0.4 = " alt="9.21 \di
Whitepunk [10]
Let us do it by converting the 0.4 to a fraction, it has one decimal, so we'll use one zero at the denominator.

and 9.21 to a fraction as well, two decimals, thus two zeros at the denominator then,

\bf 9.21\div 0.4\\\\&#10;-------------------------------\\\\&#10;0.\underline{4}\implies \cfrac{04}{1\underline{0}}\implies \cfrac{4}{10}\implies \cfrac{2}{5}\qquad \qquad \qquad \qquad 9.\underline{21}\implies \cfrac{921}{1\underline{00}}\\\\&#10;-------------------------------\\\\&#10;9.21\div 0.4\implies \cfrac{921}{100}\div \cfrac{2}{5}\implies \cfrac{921}{100}\cdot \cfrac{5}{2}\implies \cfrac{921}{20}\cdot \cfrac{1}{2}\implies \cfrac{921}{40}&#10;\\\\\\&#10;23\frac{1}{40}\implies 23.025
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