BA*BB=ABBA

Logic Level 3

$\begin{array} { l l l l } & & B & A \\ & \times & B & B \\ \hline A & B & B &A \\ \end{array}$

What is the value of the 2-digit integer, $\overline{BA}$ ?

The answer is 75.

2 solutions

Naren Bhandari
May 15, 2018

Note down the given cryptogram as $(10b+a)(11b) =10^3a +10^2b +10b +a$ We notice that left hand side is the multiple of $11$ then right hand side must be the multiple of $11$ too. Therefore using divisibility rule of $11$ we get

$a- b +b -a \equiv 0\mod(11) \\ 1001a + 110b \equiv 0\mod(11) =91a +10b \\ 91a + 10b = (10b +a)b \implies \dfrac{91a}{b} = 10b+a -10 \\ \dfrac{a}{b} = \dfrac{10b+a-10}{13\cdot 7}=\dfrac{1}{7}\left(\dfrac{10b+a-10}{13}\right) \\$ Here $a$ and $b$ are $<10$ then $b\neq 13$ . So $b =7$ and $a =5$ _ Therefore , two digit number $\overline{ba} =\boxed{75}$

I do not understand. With $x(y-1) =\equiv 0 \pmod{10}$ , we could have $y = 1, x = 3$ . Neither of these are equal to 5. Even if one of them must be 5, we could have $y -1 = 5$ right?

What do you mean by "consecutively coprime"? I've not heard that term before.

Chung Kevin - 3 years ago

Sorry Sir it was typo . I was trying to write consecutive integers . I have changed my solution . Apart from that I try to extended my solution as follows which is Incomplete solution

Incomplete solution $10b^2 +(a-10)b -91a =0$ Since the equation we have is quadratic in $b$ and solving $b$ we get $b = \dfrac{-(a-10)\pm\sqrt{(a-10)^2 + 4\cdot 10\cdot (91a)}}{20} = \dfrac{-(a-10) + \sqrt{(a-10)^2 + 4\cdot 10\cdot (91a)}}{20}$ Since $b\in\mathbb Z^+ <10$ implying that right hand side must be $<200$ also discriminant $\Delta$ should be perfect square .

Let $\Delta = p^2 = (a-10)^2 +3640a = a^2 -20a +3640a +100 \\ (p-10)(p+10) = a(a+2.1810)$

How do I solve further can you help me ? :) :). I tried however, dealing with two unknowns variables is a bit tough job and feeling lazy .

Naren Bhandari - 3 years ago

Your current solution works, but the last line could be explained better. There are a few steps skipped there.

Note that the "using divisibility rule" and subsequent comments are redundant / wrong. (For example, the second time that you use "mod 11", what you actually mean is "divide by 11" as opposed to "take the remainder after dividing by 11".) What you really want is just to divide everything by 11.

Chung Kevin - 2 years, 6 months ago

Giorgos K.
May 15, 2018

using $Mathematica$

Select[Range[11,99],#*FromDigits[{s=IntegerDigits[#][[1]],s}]==FromDigits[{t=IntegerDigits[#][[2]],s,s,t}]&]

returns $\boxed{75}$

BA*BB=ABBA

The answer is 75.

2 solutions

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