A look at two contrasting T20s, one featuring Babar Azam and one featuring Virat Kohli, a mathematical teaser and a lot of photographs.
There was much wailing and gnashing of Indian teeth this morning as the new ODI batting rankings came out with Babar Azam promoted to no1, pushing Virat Kohli down to no2. Both were in T20 action today, Babar for Pakistan against South Africa and Virat for Royal Challengers Bangalore against Sunrisers Hyderabad. This post tells the story of the international match and where we are at so far in the IPL game.
RUNS GALORE AT JO’BURG
Johannesburg is no stranger to high scoring matches (just ask Ricky Ponting, who once failed to defend 434 in an ODI there!) but even so South Africa would have expected a tally of 203 from their 20 overs to be chased down with quite such ridiculous ease. Babar Azam and Mohammad Rizwan opened the batting together and for a long time it looked like they were leading their side to a ten wicket win. Babar Azam took just 49 balls to reach his 100, and Rizwan also topped 50 quite comfortably. So unfortunately for him did Beuran Hendricks with the ball – 4-0-55-0. Eventually Babar Azam fell to the fourth ball iof the 18th over to make it 197-1, his own share 122 off 59 balls. Fakhar Zaman came in to bat and clouted the last two balls of the 18th over for fours to settle the issue with nine wickets and two whole overs unused.
RCB V SRH
Kohli was named to no one’s surprise as captain and opening batter in the Royal Challengers Bangalore XI to face Sunrisers Hyderabad. Such is Kohli’s power in certain circles that an innings of 33 off 29 balls, in reality an awful performance in a T20, was described by at least one commentator as “An excellent cameo.” Only Maxwell, who came close to living up to his moniker of “The Big Show” with 59 not out off 41 balls, did anything significant with the bat and RCB were held to 149-8 from their 20 overs, a total that seems modest. Rashid Khan as so often in any game of which is part was well to the fore with the ball, finishing with 2-18 from his four overs, and outstanding effort in this form of cricket. Although Saha fell for just one in the reply David Warner and Manish Pandey seem to be in little trouble, with SRH now 32-1 off four overs and looking set for a comfortable win.
A MATHEMATICAL TEASER
This is today’s offering from brilliant.org, slightly modified as their setting gave multiple choice options for the answer, which opened up a hack that I availed myself of. Can you solve this in the intended way and work out the answer? My hack, and an authentic solution will appear in my next post. Click here for more.
My usual sign off, with Warner and Pandey still going nicely, and Bairstow waiting to come in next…
PS as I publish, SRH are 75-1 in the tenth, well on course to chase down the modest target they have been set.
An account of today’s #INDvENG T20I cricket match and a solution to yesterday’s mathematical teaser, plus some photographs.
This post looks at an extraordinary game of cricket that has just taken place in India. I also provide a solution to the mathematical conundrum from brilliant.org that I posed yesterday and of course a few photographs.
England were unchanged, India had two changes. Ishan Kishan had a minor injury and was replaced by Suryakumar Yadav. Yuzvendra Chahal was dropped and replaced by another leg spinner, Rahul Chahar. Eoin Morgan won the toss and chose to bowl.
THE INDIAN INNINGS
Neither of India’s openers were massively convincing, and Kohli at no 4 also failed with the bat. However, Suryakumar Yadav played a quite magnificent innings, at one stage threatening to record a century, and Rishabh Pant also played very nicely. India put up 185-8 in the end, a total that looked defensible but not unassailable. Jofra Archer took four wickets, Mark Wood was also impressive, but Adil Rashid had an off day for once, and Jordan, Stokes and Curran were all unimpressive as well.
THE ENGLAND RESPONSE
Buttler failed, Malan got a bit of a start but did not go on, Roy reached 40 for the third time of the series and for the third time of the series got out with a seriously big score apparently beckoning. Bairstow and Stokes batted well together before Bairstow was out, and then it looked like Stokes and Morgan were taking England close. However, both fell to Thakur in consecutive deliveries at the start of the 18th. Curran and Jordan played decently for the rest of that over, but then Curran fell in the 19th. A four off the last ball of the 19th by Archer reduced the requirement to 23 off the final over. Thakur, who had put India in command with his bowling at the start of the 18th now lost his bearings and at one point the ask was down to ten off three balls, but then he regathered his nerve, and India emerged victorious by eight runs, setting up a final game decider on Saturday. Although the standard of play was high an both sides it is not really acceptable for 40 overs of cricket to occupy four and a quarter hours of playing time as happened today.
SOLUTION TO A TEASER
Yesterday I set you the following:
In total there are 512 small cubes in the structure. Of these 216 (6x6x6) are purely internal and therefore unpainted, eight are corner cubes and painted on three faces, which leaves 288 cubes painted on either one or two faces. The cubes painted on one face are those in the centre of each visible face, which number 36 on each face = 216 in total. This leaves 72 cubes painted on two faces, and 216 – 72 = 144. For a cube with side length n, there will eight corner pieces, (n-2) ^ 3 centre pieces that are thus unpainted, 6 ((n-2)^2) pieces that are painted on one face only and 12 (n-2) pieces that are painted on exactly two faces. Though these equations only start to work once n is greater than 2 – a 2 x 2 x 2 cube has eight blocks each of which are painted on three faces.