Mathematical Reasoning and Divisibility

This mathematics assignment focuses on mathematical reasoning and divisibility rules. It includes questions about proving divisibility by 3 and 9 using algebraic manipulation and understanding modular arithmetic. Students are also tasked with devising a similar trick for testing divisibility by 11 in base 10, demonstrating their ability to apply concepts and develop proofs. The assignment emphasizes logical thinking and problem-solving skills within the context of mathematical concepts.

Running head: MATHEMATICAL REASONING
Mathematical Reasoning
Name
Institutional Affiliation

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MATHEMATICAL REASONING
Question 1
Suppose n=¿. Show that 3∨n ⟺3∨ ( dk + dk−1+ …+d1+d0 ) .For which natural number ¿ 1
other than 3 does this trick work?
Proof:
We first need to show that indeed the trick works for n=¿
Suppose that 3∨n, where n=¿, and expressing ¿
With this notation ¿ and taking base 10 into consideration ¿, 10i can be expressed as ¿ that is
10i=¿. Hence
dk x 10k+ dk−1 x 10k−1+ dk−2 x 10k−2+ dk−3 x 10k−3+ …+d3 x 103 +d2 x 102 +d1 x 101 +d0=¿
dk x ¿
¿ ¿
{¿ ¿
{¿ ¿
Since 3∨¿. Therefore
3∨n−{¿ ¿. {HINT: if 3|n and 3|k, then 3|n-k}
But n={¿ ¿
Implying3∨{¿ ¿

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MATHEMATICAL REASONING
That is 3∨ {dk +dk−1 +dk−2 +dk−3 +… d3 +d2 +d1 +d0 }
Conversely, suppose that3∨ {dk +dk−1 +dk−2 +dk−3 +… d3 +d2 +d1 +d0 }. Then 3∨{¿ ¿. That is 3∨¿n
HINT: if 3|q and 3|k, then 3|q+k
It can easily be seen the same trick above works when 3 is replaced by 9. That is
Since 9∨¿. Therefore
9∨n−{¿ ¿.
But n={¿ ¿
Implying9∨{¿ ¿
That is 9∨ {dk +dk −1 +dk −2 +dk −3 + … d3 + d2 + d1 +d0 }
Conversely, suppose that9∨ {dk +dk −1 +dk −2 +dk −3 +… d3 + d2 + d1 +d0 }. Then 9∨{¿ ¿. That is 9∨n
Thus the same trick works for 9
Question 2

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MATHEMATICAL REASONING
Given a base b, for which integers c is it true that whenever n=¿ ¿ We have
c∨n ⟺ c∨dk +dk−1 +…+d1 +d0?
Soln.
The integers c are 3∧9 as clearly proved in question 1.
That is 3∨n ⟺3∨dk +dk −1 +…+ d1+d0 and
9∨n ⟺ 9∨dk+ dk−1+ …+d1 +d0whenever n=¿ ¿
Question 3
Devise a similar trick for testing for divisibility by 11 in base 10. Type equation here .
As a proposition, we know that any positive natural number n is divisible by 11 iff the difference
of the sums of digits in the odd and even positions results into a number that is divisible by 11
…. *
Proof:
For a more simplified formulae, we will suppose the number n=d4 d3 d2 d1 d0
Note that 10000=9999+1
1 000=1001−1
1 00=99+1

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MATHEMATICAL REASONING
1 0=11−1
All the above numbers 9999, 1001, 99, and 11 are divisible by 11 as proven by proposition *
above
Thus we can express n=d4 d3 d2 d1 d0 in the form
n=11. k +d4 −d3+ d2+ d1+ d0(devised formulae)
Proof:
Suppose that 1 1∨d4 d3 d2 d1 d0 , and since 11∨11. k, we have 11∨¿, that is
11∨ ( d4−d3+ d2+ d1 + d0 )
Conversely, suppose that 11∨ ( d4−d3+ d2+ d1 + d0 )
Since 11∨11. kit implies 11∨11. k + ( d4−d3 + d2+d1 + d0 ) that is , 11∨n

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MATHEMATICAL REASONING

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