# binary number is parallel-loaded into a shift register

A binary number is parallel-loaded into a shift register. The shift register is then commanded to shift right" for one clock pulse. How does the value of the shifted binary number compare to the number originally loaded in, assuming that the MSB is on the left flip-flop of the shift register?

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I smell homework. – Rocketmagnet Sep 4 '14 at 10:43
Just write the binary number on a piece of paper ... then do the shift and write it on the line below ... calculate the number and write next to the new binary number ..... Do this for 10 numbers between 0 and 255 (any number ... go on pick them at random if you can't think of good numbers) ... examine the pattern ... Even if you get the answer wrong you should get you a few marks.... – Spoon Sep 4 '14 at 12:31

Just try it with a couple of numbers and you can find out by yourself:
e.g. 12 (dec) has binary representation 1100.
Shift right by one bit yields 0110, which is 6 (dec).

Now try some more examples and you will see.

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A logical shift right divides the original value by 2.

Correspondingly, a logical shift left multiplies by 2.

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Ok, what did I miss? If you're going to down vote, at least explain why. – Eight-Bit Guru Sep 9 '14 at 0:52