When I get multiplied by any number, the sum of the figures in the product is always me. What am I?
2 answers:
I am the number '9'
To prove:
On multiplying 9 with different numbers, the sum of the digits of its product is always 9. Let's see;
9 * 2 = 18
where (1 + 8 = 9)
Now, multiplying it by 3
9 * 3 = 27
where (2 + 7 = 9)
Similarly,
9 * 4 = 36
where (3 + 6 = 9)
9 * 5 = 45
where (4 + 5 = 9)
9 * 6 = 54
where (5 + 4 = 9)
9 * 7 = 63
where (6 + 3 = 9).
Thus, option b i.e. 9 is the correct answer.
Learn more about 'numbers' here:
brainly.in/question/24738334
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Answer:
<u><em>The satisfied table of the given function</em></u><u><em></em></u>
<em>x 1/8 1/4 1/2 1 2</em>
<em>y -3 -2 -1 0 1</em>
<em></em>
Step-by-step explanation:
<u><em>Explanation</em></u> :-
Given logarithmic function if b >1
Given first table
i)
put x = given b > 1 so we can choose b = 2
we will apply logarithmic formula
log x ⁿ = n log (x)
<em>y = -3</em>
<em>ii)</em>
<em>put x = </em><em> given b > 1 so we can choose b = 2</em>
<em></em><em></em>
<em></em><em></em>
we will apply logarithmic formula
log x ⁿ = n log (x)
<em>y = -2</em>
<em>iii) </em>
<em>put x = </em><em> given b > 1 so we can choose b = 2</em>
<em></em><em></em>
<em>we will apply logarithmic formula </em>
<em>log x ⁿ = n log (x)</em>
<em>y = -1</em>
<em>iv) </em>
<em>put x = 1 given b > 1 so we can choose b = 2</em>
<em></em><em> = 0</em>
<em>y = 0</em>
<em>v) </em>
<em>put x = </em><em> given b > 1 so we can choose b = 2</em>
<em>y = 1</em>
<em></em>
<u><em>Final answer:-</em></u>
<u><em>The satisfied table of the given function</em></u>
<em>x 1/8 1/4 1/2 1 2</em>
<em>y -3 -2 -1 0 1</em>
<em></em>