Answers to Silly puzzles to confound you

Silly Puzzles
S1. Twenty times. The digit 5 appears ten times as the last digit (5, 15, 25 … 95) and ten times as a first digit (50, 51, 52 … 59).
S2. Five. Four daughters and a son.
S3. Think of a clock. Five hours after 9:00 AM, it is 2:00 PM!

Visual Puzzles
V1. Show fig.
V2. The first tank to fill up will be the third one, then the fourth and further the second and the first.
V3. Answer: 840. Each number is found by multiplying the number that corresponds to its step by the number that precedes it.
V4. How many circles contain a black dot? 12.

Paper and Pen Puzzles
P1. Call the numbers x, y, z, and w. So the equations shown can be written as x+y=8; z-w=6; x+z=13,y+w=8. Number these equations (1), (2), (3), (4). Add equations (2) and (4) to get z+y=14. Call this equation (5). Subtract eq. (5) from eq. (1) to get z-x=6 (call this eq. 6). But eq. (3) says x+z=13. Add and subtract these two equations to get 2z=19 and 2x=7. Hence z=19/2, x=7/2, from which we can use any suitable equation to get y=9/2 and w=7/2.

P2. This needs patience. First of all, when you multiply the rightmost number (E) by 4, the outermost number is written down, and the remaining are carried over. For instance, if E=8, then Ex4=32, and we will write down 2 and carry over 3, which we call the remainder R. So when you multiply the next number D by 4, the answer to be written down is Dx4+R. So if D=7, then we will get 28+3=31, and we will write down 1 as the next number of the answer and call 3 the new remainder.
Keeping this method in mind, let us attempt the answer. Notice that the number ABCDE multiplied by 4 gave a four digit answer. This means A is small enough so that when it is multiplied by 4 (and so remainder from the previous multiplication is added to it), it remains a single digit with no carry-over. This means that A=1 or 2 only. (If A=3, for instance, then 3x4=12, and the answer would have been 6 digits).
Now let's look at the last multiplication. Let R be the remainder when D is multiplied by 4, so that the number to be written down when A is multiplied by 4 is Ax4+R. We see from the equation ABCDE x 4=DCBA that Ax4+R must be E. Now, since we are multiplying by 4, the remainder can be R=0,1,2,3 only. Since A=1 or 2 only, Ax4+R can be 4,5,6,7,8,9. That is still a lot of numbers, but still we now know that E can be 4,5,6,7,8 or 9.
Now look at the first multiplication, starting from the right, as usual. We have Ex4=RA, where R is the remainder for this particular multiplication. For instance, if E=8, then R would have been 3. If you mentally work through your 4 times tables, you will see that A cannot be 1. So A must be 2. We have our first answer, A=2.
So Ex4 = RA, where R=0,1,2,3. Since A=2, we have Ex4=R2. Running though the 4 times tables, you can see that E=3 or 8 (4x3=12, and 4x8=32, both of which have 2 in the last place with remainder R=1 or 3).
This brings us to the next steps: Dx4+R=PB, where P is the particular remainder for this case. Also, Cx4+P=YC, where Y is the particular remainder here. Similarly, Bx4+Y=ZD. Finally, we have Ax4+Z=E, with NO REMAINDER.
At this point, we can be smart and look at the left-most multiplication again: Ax4+Z=E. We already know that A=2 and E can be 3 or 8 only. Hence, 2x4+Z=8+Z=E which is 3 or 8, or Z=0 and E=8 is the only solution. So we have the next answer: E=8. So, in the right-most multiplication, Ex4=RA, we can now solve R=3.
Let us go back to the earlier steps. Since Z=0, Bx4+Y=D. We also have Dx4+R=PB. Since R=3, we have Dx4+3=PB. Since Dx4 is an even number and 3 is an odd number, we immediately see that B is odd. Now, in Bx4+Y=D, Y is a remainder which can be 0,1,2,3. We can immediately see that B=1 or 2, since if B=3 or larger, there would be a remainder but we have already shown this is zero (Z=0). Since B is odd, this means B=1. So we have one more answer.
With B=1, the possible values of D are 4,5,6,7 (for Y=0,1,2,3). We now look at the next term: Dx4+R=PB; since R=3 and B=1, we have Dx4+3=P1, where again P=0,1,2,3. That is, we should choose one of the values of D=4,5,6,7 so that Dx4+3 gives a number ending in 1. Again, going through the 4 times tables, it is clear that D=7 alone satisfies this: 7x4+3=31. This gives us B=1 (as required) and P=3. And we have one more answer: D=7. We are almost there. Only C is to be found.
We now use this value of P to solve Cx4+P=YC. Recall that Bx4+Y=ZD. We have B=1, Z=0, D=7, so Y=3. So So, Cx4+3=3C. Since the right hand side is greater than 30, start by guessing C=7. Go through the possible values C=7,8,9 and find the equation is satisfied only when C=9: 4x9+3=39.
Hence our required number ABCDE=21978, and 21978x4=87912!

From https://www.mentalup.co/blog/brain-teasers-2