Prove that the sum of ab and ba is divisible by the sum of a and b
1 answer:
Answer:
It is divisible by 11 and (a + b) !
Step-by-step explanation:
Given a two digit number , the digits written in reverse order is
.
Note that a two digit number ab = 10a + b.
For example: 24 = 10(2) + 4
Similarly, ba = 10(b) + a
Now, the sum of the numbers ab and ba = 10a + b + 10b + a
= 11a + 11b
= 11(a + b)
Hence, the sum of any two digit number ab and the reverse of the number ba, is divisible by 11 and (a + b).
Hence, proved.
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Answer:
A dilation by a factor of three about Point T followed by a translation of two units downwards.
Step-by-step explanation:
When transforming functions, we will reflect/dilate the figure first and then translate it. This is directly from the order of operations.
Since we are trying to determine the transformation that was performed, we can try to map ΔS'T'U' onto ΔSTU. We can start by translating the figure and then determining any reflections/dilations.
First, we can translate ΔS'T'U' up two units to map T' onto T. This is represented by the black triangle in the image below. Let the black triangle be ΔS''T''U''. (T'' and T are the same point.)
Next, notice that from Point T'' to U'', we move nine units right and six units up.
From Point T to Point U, we move three units right and two units up.
Likewise, from Point T'' to S'', we move six units left and nine units up.
From Point T to Point S, we move two units left and three units up.
Therefore, to map ΔS''T''U'' onto ΔSTU, we dilate ΔS''T''U'' about Point T by a factor of 1/3.
Hence, by reversing the transformations, to acquire ΔS'T'U', we can see that we will dilate ΔSTU by a factor of three about Point T and then a perform a translation of two units downwards.
